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Question 9 1pts1 \mathrm{pts} Consider a firm with a production function given by...
May 15, 2024
Solution by Steps
step 1
The production function is given by F(L,K)=L+K+ln⁡(LK)F(L, K) = L + K + \ln(LK)
step 2
To find the marginal rate of technical substitution (MRTS), we need to compute the partial derivatives of FF with respect to LL and KK
step 3
The partial derivative of FF with respect to LL is ∂F∂L=1+1L\frac{\partial F}{\partial L} = 1 + \frac{1}{L}
step 4
The partial derivative of FF with respect to KK is ∂F∂K=1+1K\frac{\partial F}{\partial K} = 1 + \frac{1}{K}
step 5
The MRTS is given by the negative ratio of these partial derivatives: MRTS=−∂F∂L∂F∂K\text{MRTS} = -\frac{\frac{\partial F}{\partial L}}{\frac{\partial F}{\partial K}}
step 6
At the point (4,1)(4,1), we substitute L=4L = 4 and K=1K = 1 into the partial derivatives: ∂F∂L=1+14=1.25\frac{\partial F}{\partial L} = 1 + \frac{1}{4} = 1.25 and ∂F∂K=1+11=2\frac{\partial F}{\partial K} = 1 + \frac{1}{1} = 2
step 7
Therefore, the MRTS at the point (4,1)(4,1) is MRTS=−1.252=−0.625\text{MRTS} = -\frac{1.25}{2} = -0.625
Answer
The marginal rate of technical substitution at the point (4,1)(4,1) is −0.625-0.625.
Key Concept
Marginal Rate of Technical Substitution (MRTS)
Explanation
The MRTS is the rate at which one input can be substituted for another input while keeping the output level constant. It is calculated as the negative ratio of the marginal products of the inputs.
Solution by Steps
step 1
The utility function is given by U(x1,x2)=x12x22+x1+x2U(x_1, x_2) = x_1^2 x_2^2 + x_1 + x_2
step 2
To find the marginal utility with respect to x1x_1, we compute the partial derivative of UU with respect to x1x_1: ∂U∂x1=2x1x22+1 \frac{\partial U}{\partial x_1} = 2x_1 x_2^2 + 1
step 3
We are given that the marginal utility with respect to x1x_1 at the point (x1,3)(x_1, 3) is 10. Therefore, we substitute x2=3x_2 = 3 and set the partial derivative equal to 10: 2x1(3)2+1=10 2x_1 (3)^2 + 1 = 10
step 4
Simplifying the equation: 18x1+1=10 18x_1 + 1 = 10 18x1=9 18x_1 = 9 x1=918=12 x_1 = \frac{9}{18} = \frac{1}{2}
Answer
x1=12x_1 = \frac{1}{2}
Key Concept
Marginal Utility Calculation
Explanation
The marginal utility with respect to a variable is found by taking the partial derivative of the utility function with respect to that variable and then solving for the given conditions.
Question 11
step 1
The utility function is given by U(x1,x2)=x12x22+x1+x2U(x_1, x_2) = x_1^2 x_2^2 + x_1 + x_2
step 2
To find the marginal utility with respect to x2x_2, we compute the partial derivative of UU with respect to x2x_2: ∂U∂x2=2x12x2+1 \frac{\partial U}{\partial x_2} = 2x_1^2 x_2 + 1
step 3
We are given that the marginal utility with respect to x2x_2 at the point (5,x2)(5, x_2) is 5. Therefore, we substitute x1=5x_1 = 5 and set the partial derivative equal to 5: 2(5)2x2+1=5 2(5)^2 x_2 + 1 = 5
step 4
Simplifying the equation: 50x2+1=5 50x_2 + 1 = 5 50x2=4 50x_2 = 4 x2=450=225 x_2 = \frac{4}{50} = \frac{2}{25}
Answer
x2=225x_2 = \frac{2}{25}
Key Concept
Marginal Utility Calculation
Explanation
The marginal utility with respect to a variable is found by taking the partial derivative of the utility function with respect to that variable and then solving for the given conditions.
Solution by Steps
step 1
The marginal cost function is given by MC=−Q2+80QMC = -Q^2 + 80Q
step 2
The total cost function C(Q)C(Q) is related to the marginal cost by the differential equation dCdQ=MC\frac{dC}{dQ} = MC
step 3
Therefore, the differential equation is dCdQ=−Q2+80Q\frac{dC}{dQ} = -Q^2 + 80Q
# (b) Find the equation of the total cost function if fixed costs are 500.
step 1
Integrate the marginal cost function to find the total cost function: ∫(−Q2+80Q) dQ\int (-Q^2 + 80Q) \, dQ
step 2
The integral of −Q2-Q^2 is −Q33-\frac{Q^3}{3} and the integral of 80Q80Q is 40Q240Q^2
step 3
Therefore, C(Q)=−Q33+40Q2+C0C(Q) = -\frac{Q^3}{3} + 40Q^2 + C_0, where C0C_0 is the constant of integration
step 4
Given that the fixed costs are 500, we set C(0)=500C(0) = 500
step 5
Thus, C0=500C_0 = 500
step 6
The total cost function is C(Q)=−Q33+40Q2+500C(Q) = -\frac{Q^3}{3} + 40Q^2 + 500
# (c) Calculate the cost of producing successive units from Q=3Q=3 to Q=12Q=12.
step 1
Use the total cost function C(Q)=−Q33+40Q2+500C(Q) = -\frac{Q^3}{3} + 40Q^2 + 500
step 2
Calculate C(3)C(3): C(3)=−333+40⋅32+500=−9+360+500=851C(3) = -\frac{3^3}{3} + 40 \cdot 3^2 + 500 = -9 + 360 + 500 = 851
step 3
Calculate C(12)C(12): C(12)=−1233+40⋅122+500=−576+5760+500=5684C(12) = -\frac{12^3}{3} + 40 \cdot 12^2 + 500 = -576 + 5760 + 500 = 5684
step 4
The cost of producing units from Q=3Q=3 to Q=12Q=12 is C(12)−C(3)=5684−851=4833C(12) - C(3) = 5684 - 851 = 4833
Answer
The cost of producing units from Q=3Q=3 to Q=12Q=12 is 4833.
Question 2 # (a) Find the equation of the total cost function if fixed costs are 900.
step 1
The marginal cost function is given by MC=Q2−42Q+200MC = Q^2 - 42Q + 200
step 2
The total cost function C(Q)C(Q) is related to the marginal cost by the differential equation dCdQ=MC\frac{dC}{dQ} = MC
step 3
Therefore, the differential equation is dCdQ=Q2−42Q+200\frac{dC}{dQ} = Q^2 - 42Q + 200
step 4
Integrate the marginal cost function to find the total cost function: ∫(Q2−42Q+200) dQ\int (Q^2 - 42Q + 200) \, dQ
step 5
The integral of Q2Q^2 is Q33\frac{Q^3}{3}, the integral of −42Q-42Q is −21Q2-21Q^2, and the integral of 200200 is 200Q200Q
step 6
Therefore, C(Q)=Q33−21Q2+200Q+C0C(Q) = \frac{Q^3}{3} - 21Q^2 + 200Q + C_0, where C0C_0 is the constant of integration
step 7
Given that the fixed costs are 900, we set C(0)=900C(0) = 900
step 8
Thus, C0=900C_0 = 900
step 9
The total cost function is C(Q)=Q33−21Q2+200Q+900C(Q) = \frac{Q^3}{3} - 21Q^2 + 200Q + 900
# (b) Calculate the total cost of producing 12 units.
step 1
Use the total cost function C(Q)=Q33−21Q2+200Q+900C(Q) = \frac{Q^3}{3} - 21Q^2 + 200Q + 900
step 2
Calculate C(12)C(12): C(12)=1233−21⋅122+200⋅12+900C(12) = \frac{12^3}{3} - 21 \cdot 12^2 + 200 \cdot 12 + 900
step 3
Simplify: C(12)=17283−3024+2400+900=576−3024+2400+900=852C(12) = \frac{1728}{3} - 3024 + 2400 + 900 = 576 - 3024 + 2400 + 900 = 852
Answer
The total cost of producing 12 units is 852.
Key Concept
Marginal Cost and Total Cost Functions
Explanation
The marginal cost function represents the cost of producing one more unit of a good. By integrating the marginal cost function and adding fixed costs, we can find the total cost function.
Solution by Steps
step 1
The given differential equation is dPdt=0.01P\frac{dP}{dt} = 0.01P. This is a first-order linear differential equation
step 2
To solve it, we separate the variables: dPP=0.01 dt\frac{dP}{P} = 0.01 \, dt
step 3
Integrate both sides: ∫1P dP=∫0.01 dt\int \frac{1}{P} \, dP = \int 0.01 \, dt
step 4
This gives ln⁡P=0.01t+C\ln P = 0.01t + C, where CC is the integration constant
step 5
Exponentiate both sides to solve for PP: P=e0.01t+C=eC⋅e0.01tP = e^{0.01t + C} = e^C \cdot e^{0.01t}
step 6
Let eC=P0e^C = P_0, the initial population. Thus, P=P0e0.01tP = P_0 e^{0.01t}
Answer
P=P0e0.01tP = P_0 e^{0.01t}
Question 2(b)
step 1
Given P=58.6P = 58.6 million in 1998, let t=0t = 0 at the start of 1998. Thus, P0=58.6P_0 = 58.6
step 2
We need to find the population in the year 2200. The time difference from 1998 to 2200 is t=2200−1998=202t = 2200 - 1998 = 202 years
step 3
Substitute P0=58.6P_0 = 58.6 and t=202t = 202 into the equation P=P0e0.01tP = P_0 e^{0.01t}:
step 4
P=58.6⋅e0.01⋅202P = 58.6 \cdot e^{0.01 \cdot 202}
step 5
Calculate e0.01⋅202≈e2.02≈7.53e^{0.01 \cdot 202} \approx e^{2.02} \approx 7.53
step 6
Thus, P≈58.6⋅7.53≈441.258P \approx 58.6 \cdot 7.53 \approx 441.258 million
Answer
P≈441.258P \approx 441.258 million
Question 2(c)
step 1
We need to find the time tt when the population reaches 70 million
step 2
Use the equation P=P0e0.01tP = P_0 e^{0.01t} with P=70P = 70 and P0=58.6P_0 = 58.6:
step 3
70=58.6⋅e0.01t70 = 58.6 \cdot e^{0.01t}
step 4
Divide both sides by 58.6: 7058.6=e0.01t\frac{70}{58.6} = e^{0.01t}
step 5
Calculate 7058.6≈1.195\frac{70}{58.6} \approx 1.195
step 6
Take the natural logarithm of both sides: ln⁡(1.195)=0.01t\ln(1.195) = 0.01t
step 7
Solve for tt: t=ln⁡(1.195)0.01≈0.1780.01≈17.8t = \frac{\ln(1.195)}{0.01} \approx \frac{0.178}{0.01} \approx 17.8 years
Answer
t≈17.8t \approx 17.8 years
Question 4(a)
step 1
The given differential equation is dIdt=−0.05I\frac{dI}{dt} = -0.05I
step 2
Separate the variables: dII=−0.05 dt\frac{dI}{I} = -0.05 \, dt
step 3
Integrate both sides: ∫1I dI=∫−0.05 dt\int \frac{1}{I} \, dI = \int -0.05 \, dt
step 4
This gives ln⁡I=−0.05t+C\ln I = -0.05t + C, where CC is the integration constant
step 5
Exponentiate both sides to solve for II: I=e−0.05t+C=eC⋅e−0.05tI = e^{-0.05t + C} = e^C \cdot e^{-0.05t}
step 6
Let eC=I0e^C = I_0, the initial investment. Thus, I=I0e−0.05tI = I_0 e^{-0.05t}
step 7
Given I0=£12000I_0 = £12000 and t=5.5t = 5.5 years, substitute these values into the equation:
step 8
I=12000⋅e−0.05⋅5.5I = 12000 \cdot e^{-0.05 \cdot 5.5}
step 9
Calculate e−0.275≈0.759e^{-0.275} \approx 0.759
step 10
Thus, I≈12000⋅0.759≈9108I \approx 12000 \cdot 0.759 \approx 9108
Answer
£9108
Question 4(b)
step 1
We need to find the time tt when the investment value falls to £5000
step 2
Use the equation I=I0e−0.05tI = I_0 e^{-0.05t} with I=5000I = 5000 and I0=12000I_0 = 12000:
step 3
5000=12000⋅e−0.05t5000 = 12000 \cdot e^{-0.05t}
step 4
Divide both sides by 12000: 500012000=e−0.05t\frac{5000}{12000} = e^{-0.05t}
step 5
Calculate 500012000≈0.4167\frac{5000}{12000} \approx 0.4167
step 6
Take the natural logarithm of both sides: ln⁡(0.4167)=−0.05t\ln(0.4167) = -0.05t
step 7
Solve for tt: t=ln⁡(0.4167)−0.05≈−0.875−0.05≈17.5t = \frac{\ln(0.4167)}{-0.05} \approx \frac{-0.875}{-0.05} \approx 17.5 years
Answer
17.5 years
Key Concept
Differential Equations in Population Growth and Depreciation
Explanation
The problems involve solving first-order linear differential equations to model population growth and investment depreciation. The solutions require separating variables, integrating, and applying initial conditions.
matrics
Solution by Steps
step 1
To find the marginal rate of technical substitution (MRTS), we need to understand the firm's production function. The MRTS is the rate at which one input can be substituted for another while keeping the output constant
step 2
The MRTS is calculated as the negative ratio of the marginal products of the inputs. If the production function is Q=f(K,L)Q = f(K, L), where KK is capital and LL is labor, then the MRTS is given by: MRTS=−∂Q/∂L∂Q/∂K \text{MRTS} = -\frac{\partial Q / \partial L}{\partial Q / \partial K}
step 3
To find the marginal products, we take the partial derivatives of the production function with respect to each input. For example, if the production function is Q=K0.5L0.5Q = K^{0.5}L^{0.5}, then: ∂Q∂L=0.5K0.5L−0.5 \frac{\partial Q}{\partial L} = 0.5 K^{0.5} L^{-0.5} ∂Q∂K=0.5K−0.5L0.5 \frac{\partial Q}{\partial K} = 0.5 K^{-0.5} L^{0.5}
step 4
Substituting these into the MRTS formula, we get: MRTS=−0.5K0.5L−0.50.5K−0.5L0.5=−KL \text{MRTS} = -\frac{0.5 K^{0.5} L^{-0.5}}{0.5 K^{-0.5} L^{0.5}} = -\frac{K}{L}
step 5
To find the MRTS at a specific point, substitute the values of KK and LL at that point. For example, if K=4K = 4 and L=2L = 2, then: MRTS=−42=−2 \text{MRTS} = -\frac{4}{2} = -2
Answer
The marginal rate of technical substitution (MRTS) at the given point is -2.
Key Concept
Marginal Rate of Technical Substitution (MRTS)
Explanation
The MRTS measures the rate at which one input can be substituted for another while keeping the output constant. It is calculated as the negative ratio of the marginal products of the inputs.
Solution by Steps
step 1
To find the transpose of matrix AA, we need to switch its rows and columns. Matrix AA is given as: A=(1amp;5amp;64amp;7amp;9) A = \begin{pmatrix} 1 & 5 & 6 \\ 4 & 7 & 9 \end{pmatrix}
step 2
The transpose of matrix AA, denoted as ATA^T, is obtained by switching the rows and columns: AT=(1amp;45amp;76amp;9) A^T = \begin{pmatrix} 1 & 4 \\ 5 & 7 \\ 6 & 9 \end{pmatrix}
step 3
To find the transpose of matrix BB, we need to switch its rows and columns. Matrix BB is given as: B=(4amp;4−1amp;32amp;13amp;3) B = \begin{pmatrix} 4 & 4 \\ -1 & 3 \\ 2 & 1 \\ 3 & 3 \end{pmatrix}
step 4
The transpose of matrix BB, denoted as BTB^T, is obtained by switching the rows and columns: BT=(4amp;−1amp;2amp;34amp;3amp;1amp;3) B^T = \begin{pmatrix} 4 & -1 & 2 & 3 \\ 4 & 3 & 1 & 3 \end{pmatrix}
Answer
The transpose of matrix AA is: AT=(1amp;45amp;76amp;9) A^T = \begin{pmatrix} 1 & 4 \\ 5 & 7 \\ 6 & 9 \end{pmatrix} The transpose of matrix BB is: BT=(4amp;−1amp;2amp;34amp;3amp;1amp;3) B^T = \begin{pmatrix} 4 & -1 & 2 & 3 \\ 4 & 3 & 1 & 3 \end{pmatrix}
Key Concept
Transpose of a Matrix
Explanation
The transpose of a matrix is obtained by switching its rows and columns. This means that the element at the ii-th row and jj-th column of the original matrix becomes the element at the jj-th row and ii-th column of the transposed matrix.
Solution by Steps
step 1
To find A+BA + B, we add the corresponding elements of matrices AA and BB
step 2
A=(2amp;30amp;3−2amp;6)A = \left(\begin{array}{cc}2 & 3 \\ 0 & 3 \\ -2 & 6\end{array}\right) and B=(1amp;7−1amp;42amp;2)B = \left(\begin{array}{cc}1 & 7 \\ -1 & 4 \\ 2 & 2\end{array}\right)
step 3
Adding the corresponding elements: A+B=(2+1amp;3+70+(−1)amp;3+4−2+2amp;6+2)=(3amp;10−1amp;70amp;8) A + B = \left(\begin{array}{cc}2+1 & 3+7 \\ 0+(-1) & 3+4 \\ -2+2 & 6+2\end{array}\right) = \left(\begin{array}{cc}3 & 10 \\ -1 & 7 \\ 0 & 8\end{array}\right)
Answer
A+B=(3amp;10−1amp;70amp;8)A + B = \left(\begin{array}{cc}3 & 10 \\ -1 & 7 \\ 0 & 8\end{array}\right)
step 1
To find A−BA - B, we subtract the corresponding elements of matrix BB from matrix AA
step 2
A=(2amp;30amp;3−2amp;6)A = \left(\begin{array}{cc}2 & 3 \\ 0 & 3 \\ -2 & 6\end{array}\right) and B=(1amp;7−1amp;42amp;2)B = \left(\begin{array}{cc}1 & 7 \\ -1 & 4 \\ 2 & 2\end{array}\right)
step 3
Subtracting the corresponding elements: A−B=(2−1amp;3−70−(−1)amp;3−4−2−2amp;6−2)=(1amp;−41amp;−1−4amp;4) A - B = \left(\begin{array}{cc}2-1 & 3-7 \\ 0-(-1) & 3-4 \\ -2-2 & 6-2\end{array}\right) = \left(\begin{array}{cc}1 & -4 \\ 1 & -1 \\ -4 & 4\end{array}\right)
Answer
A−B=(1amp;−41amp;−1−4amp;4)A - B = \left(\begin{array}{cc}1 & -4 \\ 1 & -1 \\ -4 & 4\end{array}\right)
Key Concept
Matrix Addition and Subtraction
Explanation
Matrix addition and subtraction involve adding or subtracting corresponding elements from each matrix.
Solution by Steps
step 1
Given the matrix A=(2amp;30amp;3−2amp;6)A = \left(\begin{array}{cc}2 & 3 \\ 0 & 3 \\ -2 & 6\end{array}\right), we need to calculate 5A5A
step 2
To find 5A5A, we multiply each element of matrix AA by 5
step 3
Performing the multiplication: 5A=5(2amp;30amp;3−2amp;6)=(5⋅2amp;5⋅35⋅0amp;5⋅35⋅(−2)amp;5⋅6)=(10amp;150amp;15−10amp;30) 5A = 5 \left(\begin{array}{cc}2 & 3 \\ 0 & 3 \\ -2 & 6\end{array}\right) = \left(\begin{array}{cc}5 \cdot 2 & 5 \cdot 3 \\ 5 \cdot 0 & 5 \cdot 3 \\ 5 \cdot (-2) & 5 \cdot 6\end{array}\right) = \left(\begin{array}{cc}10 & 15 \\ 0 & 15 \\ -10 & 30\end{array}\right)
step 4
The resulting matrix after multiplication is: 5A=(10amp;150amp;15−10amp;30) 5A = \left(\begin{array}{cc}10 & 15 \\ 0 & 15 \\ -10 & 30\end{array}\right)
Answer
5A=(10amp;150amp;15−10amp;30)5A = \left(\begin{array}{cc}10 & 15 \\ 0 & 15 \\ -10 & 30\end{array}\right)
Key Concept
Matrix Scalar Multiplication
Explanation
To multiply a matrix by a scalar, multiply each element of the matrix by the scalar.
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Solution by Steps
step 1
Given the identity matrix II of size 3×33 \times 3, we need to calculate 3I3I
step 2
The 3×33 \times 3 identity matrix II is: I=(1amp;0amp;00amp;1amp;00amp;0amp;1) I = \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)
step 3
To find 3I3I, we multiply each element of the identity matrix II by 3
step 4
Performing the multiplication: 3I=3(1amp;0amp;00amp;1amp;00amp;0amp;1)=(3⋅1amp;3⋅0amp;3⋅03⋅0amp;3⋅1amp;3⋅03⋅0amp;3⋅0amp;3⋅1)=(3amp;0amp;00amp;3amp;00amp;0amp;3) 3I = 3 \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right) = \left(\begin{array}{ccc}3 \cdot 1 & 3 \cdot 0 & 3 \cdot 0 \\ 3 \cdot 0 & 3 \cdot 1 & 3 \cdot 0 \\ 3 \cdot 0 & 3 \cdot 0 & 3 \cdot 1\end{array}\right) = \left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right)
step 5
The resulting matrix after multiplication is: 3I=(3amp;0amp;00amp;3amp;00amp;0amp;3) 3I = \left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right)
Answer
3I=(3amp;0amp;00amp;3amp;00amp;0amp;3)3I = \left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right)
Key Concept
Identity Matrix
Explanation
The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. Multiplying it by a scalar scales the diagonal elements by that scalar.
Solution by Steps
step 1
To find the product of matrices A\mathrm{A} and B\mathrm{B}, we need to perform matrix multiplication
step 2
Given A=(3amp;9amp;−1)\mathrm{A}=\left(\begin{array}{lll}3 & 9 & -1\end{array}\right) and B=(231)\mathrm{B}=\left(\begin{array}{l}2 \\ 3 \\ 1\end{array}\right), we multiply each element of A\mathrm{A} by the corresponding element of B\mathrm{B} and sum the results
step 3
The product A×B\mathrm{A} \times \mathrm{B} is calculated as follows: A×B=3⋅2+9⋅3+(−1)⋅1=6+27−1=32 \mathrm{A} \times \mathrm{B} = 3 \cdot 2 + 9 \cdot 3 + (-1) \cdot 1 = 6 + 27 - 1 = 32
step 4
For the second example, given A=(1amp;−1)\mathrm{A}=\left(\begin{array}{ll}1 & -1\end{array}\right) and B=(−11)\mathrm{B}=\binom{-1}{1}, we perform the same matrix multiplication process
step 5
The product A×B\mathrm{A} \times \mathrm{B} is calculated as follows: A×B=1⋅(−1)+(−1)⋅1=−1−1=−2 \mathrm{A} \times \mathrm{B} = 1 \cdot (-1) + (-1) \cdot 1 = -1 - 1 = -2
Answer
The product of the first pair of matrices is 3232. The product of the second pair of matrices is −2-2.
Key Concept
Matrix Multiplication
Explanation
Matrix multiplication involves multiplying corresponding elements and summing the results.
Solution by Steps
step 1
To find the product of matrices AA and CC, we need to multiply each element of the rows of AA by the corresponding elements of the columns of CC and sum the products
step 2
Matrix AA is given by: A=(1amp;2−2amp;4) A = \begin{pmatrix} 1 & 2 \\ -2 & 4 \end{pmatrix} Matrix CC is given by: C=(3amp;−25amp;0) C = \begin{pmatrix} 3 & -2 \\ 5 & 0 \end{pmatrix}
step 3
Calculate the element in the first row and first column of ACAC: (1⋅3)+(2⋅5)=3+10=13 (1 \cdot 3) + (2 \cdot 5) = 3 + 10 = 13
step 4
Calculate the element in the first row and second column of ACAC: (1⋅−2)+(2⋅0)=−2+0=−2 (1 \cdot -2) + (2 \cdot 0) = -2 + 0 = -2
step 5
Calculate the element in the second row and first column of ACAC: (−2⋅3)+(4⋅5)=−6+20=14 (-2 \cdot 3) + (4 \cdot 5) = -6 + 20 = 14
step 6
Calculate the element in the second row and second column of ACAC: (−2⋅−2)+(4⋅0)=4+0=4 (-2 \cdot -2) + (4 \cdot 0) = 4 + 0 = 4
step 7
Combine the results to form the matrix ACAC: AC=(13amp;−214amp;4) AC = \begin{pmatrix} 13 & -2 \\ 14 & 4 \end{pmatrix}
Answer
AC=(13amp;−214amp;4)AC = \begin{pmatrix} 13 & -2 \\ 14 & 4 \end{pmatrix}
Question 2: What is CA?
step 1
To find the product of matrices CC and AA, we need to multiply each element of the rows of CC by the corresponding elements of the columns of AA and sum the products
step 2
Matrix CC is given by: C=(3amp;−25amp;0) C = \begin{pmatrix} 3 & -2 \\ 5 & 0 \end{pmatrix} Matrix AA is given by: A=(1amp;2−2amp;4) A = \begin{pmatrix} 1 & 2 \\ -2 & 4 \end{pmatrix}
step 3
Calculate the element in the first row and first column of CACA: (3⋅1)+(−2⋅−2)=3+4=7 (3 \cdot 1) + (-2 \cdot -2) = 3 + 4 = 7
step 4
Calculate the element in the first row and second column of CACA: (3⋅2)+(−2⋅4)=6−8=−2 (3 \cdot 2) + (-2 \cdot 4) = 6 - 8 = -2
step 5
Calculate the element in the second row and first column of CACA: (5⋅1)+(0⋅−2)=5+0=5 (5 \cdot 1) + (0 \cdot -2) = 5 + 0 = 5
step 6
Calculate the element in the second row and second column of CACA: (5⋅2)+(0⋅4)=10+0=10 (5 \cdot 2) + (0 \cdot 4) = 10 + 0 = 10
step 7
Combine the results to form the matrix CACA: CA=(7amp;−25amp;10) CA = \begin{pmatrix} 7 & -2 \\ 5 & 10 \end{pmatrix}
Answer
CA=(7amp;−25amp;10)CA = \begin{pmatrix} 7 & -2 \\ 5 & 10 \end{pmatrix}
Question 3: What is AB?
step 1
To find the product of matrices AA and BB, we need to multiply each element of the rows of AA by the corresponding elements of the columns of BB and sum the products
step 2
Matrix AA is given by: A=(1amp;2−2amp;4) A = \begin{pmatrix} 1 & 2 \\ -2 & 4 \end{pmatrix} Matrix BB is given by: B=(0amp;2amp;21amp;0amp;5) B = \begin{pmatrix} 0 & 2 & 2 \\ 1 & 0 & 5 \end{pmatrix}
step 3
Calculate the element in the first row and first column of ABAB: (1⋅0)+(2⋅1)=0+2=2 (1 \cdot 0) + (2 \cdot 1) = 0 + 2 = 2
step 4
Calculate the element in the first row and second column of ABAB: (1⋅2)+(2⋅0)=2+0=2 (1 \cdot 2) + (2 \cdot 0) = 2 + 0 = 2
step 5
Calculate the element in the first row and third column of ABAB: (1⋅2)+(2⋅5)=2+10=12 (1 \cdot 2) + (2 \cdot 5) = 2 + 10 = 12
step 6
Calculate the element in the second row and first column of ABAB: (−2⋅0)+(4⋅1)=0+4=4 (-2 \cdot 0) + (4 \cdot 1) = 0 + 4 = 4
step 7
Calculate the element in the second row and second column of ABAB: (−2⋅2)+(4⋅0)=−4+0=−4 (-2 \cdot 2) + (4 \cdot 0) = -4 + 0 = -4
step 8
Calculate the element in the second row and third column of ABAB: (−2⋅2)+(4⋅5)=−4+20=16 (-2 \cdot 2) + (4 \cdot 5) = -4 + 20 = 16
step 9
Combine the results to form the matrix ABAB: AB=(2amp;2amp;124amp;−4amp;16) AB = \begin{pmatrix} 2 & 2 & 12 \\ 4 & -4 & 16 \end{pmatrix}
Answer
AB=(2amp;2amp;124amp;−4amp;16)AB = \begin{pmatrix} 2 & 2 & 12 \\ 4 & -4 & 16 \end{pmatrix}
Key Concept
Matrix Multiplication
Explanation
Matrix multiplication involves taking the dot product of rows and columns from the two matrices being multiplied. The resulting matrix has dimensions based on the number of rows of the first matrix and the number of columns of the second matrix.
Solution by Steps
step 1
First, we need to multiply matrices AA and BB. Matrix AA is given by: A=(5amp;1amp;02amp;6amp;−1) A = \begin{pmatrix} 5 & 1 & 0 \\ 2 & 6 & -1 \end{pmatrix} and matrix BB is given by: B=(4amp;31amp;10amp;2) B = \begin{pmatrix} 4 & 3 \\ 1 & 1 \\ 0 & 2 \end{pmatrix}
step 2
To find the product ABAB, we perform matrix multiplication. The element in the first row and first column of ABAB is calculated as: (5⋅4)+(1⋅1)+(0⋅0)=20+1+0=21 (5 \cdot 4) + (1 \cdot 1) + (0 \cdot 0) = 20 + 1 + 0 = 21
step 3
The element in the first row and second column of ABAB is calculated as: (5⋅3)+(1⋅1)+(0⋅2)=15+1+0=16 (5 \cdot 3) + (1 \cdot 1) + (0 \cdot 2) = 15 + 1 + 0 = 16
step 4
The element in the second row and first column of ABAB is calculated as: (2⋅4)+(6⋅1)+(−1⋅0)=8+6+0=14 (2 \cdot 4) + (6 \cdot 1) + (-1 \cdot 0) = 8 + 6 + 0 = 14
step 5
The element in the second row and second column of ABAB is calculated as: (2⋅3)+(6⋅1)+(−1⋅2)=6+6−2=10 (2 \cdot 3) + (6 \cdot 1) + (-1 \cdot 2) = 6 + 6 - 2 = 10
step 6
Therefore, the product ABAB is: AB=(21amp;1614amp;10) AB = \begin{pmatrix} 21 & 16 \\ 14 & 10 \end{pmatrix}
step 7
Comparing this with the given matrix: AB=(xamp;1614amp;y) AB = \begin{pmatrix} x & 16 \\ 14 & y \end{pmatrix} we can see that x=21x = 21 and y=10y = 10
Answer
x=21x = 21, y=10y = 10
Key Concept
Matrix Multiplication
Explanation
Matrix multiplication involves taking the dot product of rows and columns from the two matrices.
Solution by Steps
step 2
To calculate A−BA - B, we subtract the corresponding elements of matrices AA and BB: A−B=(1−4amp;−4−30−(−7)amp;−9−0)=(−3amp;−77amp;−9) A - B = \begin{pmatrix} 1 - 4 & -4 - 3 \\ 0 - (-7) & -9 - 0 \end{pmatrix} = \begin{pmatrix} -3 & -7 \\ 7 & -9 \end{pmatrix}
step 3
To calculate A+4BA + 4B, we first multiply matrix BB by 4 and then add the result to matrix AA: 4B=4(4amp;3−7amp;0)=(16amp;12−28amp;0) 4B = 4 \begin{pmatrix} 4 & 3 \\ -7 & 0 \end{pmatrix} = \begin{pmatrix} 16 & 12 \\ -28 & 0 \end{pmatrix} A+4B=(1amp;−40amp;−9)+(16amp;12−28amp;0)=(17amp;8−28amp;−9) A + 4B = \begin{pmatrix} 1 & -4 \\ 0 & -9 \end{pmatrix} + \begin{pmatrix} 16 & 12 \\ -28 & 0 \end{pmatrix} = \begin{pmatrix} 17 & 8 \\ -28 & -9 \end{pmatrix}
step 4
To calculate A+IA + I, where II is the identity matrix of the same size as AA: I=(1amp;00amp;1) I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} A+I=(1amp;−40amp;−9)+(1amp;00amp;1)=(2amp;−40amp;−8) A + I = \begin{pmatrix} 1 & -4 \\ 0 & -9 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & -4 \\ 0 & -8 \end{pmatrix}
step 5
To calculate AIAI, where II is the identity matrix: AI=(1amp;−40amp;−9)⋅(1amp;00amp;1)=(1amp;−40amp;−9) AI = \begin{pmatrix} 1 & -4 \\ 0 & -9 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ 0 & -9 \end{pmatrix}
step 6
To calculate A+CA + C, we note that AA is a 2×22 \times 2 matrix and CC is a 2×32 \times 3 matrix. Since the matrices have different dimensions, the operation A+CA + C is not defined
[question number] Answer
A
Key Concept
Matrix Addition and Subtraction
Explanation
Matrix addition and subtraction are performed element-wise, and the matrices must have the same dimensions.
Solution by Steps
step 2
det⁡(A)=(1⋅4)−(2⋅3)=4−6=−2\operatorname{det}(A) = (1 \cdot 4) - (2 \cdot 3) = 4 - 6 = -2
step 3
Calculate the determinant of matrix B=(3amp;46amp;8)B = \left(\begin{array}{ll}3 & 4 \\ 6 & 8\end{array}\right) using the formula det⁡(B)=ad−bc\operatorname{det}(B) = ad - bc
step 4
det⁡(B)=(3⋅8)−(4⋅6)=24−24=0\operatorname{det}(B) = (3 \cdot 8) - (4 \cdot 6) = 24 - 24 = 0
step 5
Calculate the product of matrices AA and BB: AB=(1amp;23amp;4)(3amp;46amp;8)AB = \left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right) \left(\begin{array}{ll}3 & 4 \\ 6 & 8\end{array}\right)
step 6
AB=((1⋅3+2⋅6)amp;(1⋅4+2⋅8)(3⋅3+4⋅6)amp;(3⋅4+4⋅8))=(15amp;2033amp;44)AB = \left(\begin{array}{ll}(1 \cdot 3 + 2 \cdot 6) & (1 \cdot 4 + 2 \cdot 8) \\ (3 \cdot 3 + 4 \cdot 6) & (3 \cdot 4 + 4 \cdot 8)\end{array}\right) = \left(\begin{array}{ll}15 & 20 \\ 33 & 44\end{array}\right)
step 7
Calculate the determinant of matrix ABAB: det⁡(AB)=(15⋅44)−(20⋅33)=660−660=0\operatorname{det}(AB) = (15 \cdot 44) - (20 \cdot 33) = 660 - 660 = 0
step 8
Compare det⁡(AB)\operatorname{det}(AB) with det⁡(A)⋅det⁡(B)\operatorname{det}(A) \cdot \operatorname{det}(B): 0=−2⋅0=00 = -2 \cdot 0 = 0
step 9
Since det⁡(AB)=det⁡(A)⋅det⁡(B)\operatorname{det}(AB) = \operatorname{det}(A) \cdot \operatorname{det}(B), the statement is True
True
Key Concept
Determinant of a Product of Matrices
Explanation
The determinant of the product of two matrices is equal to the product of their determinants.
Generated Graph
Solution by Steps
step 1
Given the inverse demand functions for goods XX and YY are P=24−xP = 24 - x and P=32−yP = 32 - y respectively. To find the optimal levels of xx and yy, we need to maximize the revenue for each good
step 2
The revenue for good XX is RX=PX⋅x=(24−x)⋅xR_X = P_X \cdot x = (24 - x) \cdot x
step 3
To find the maximum revenue, we take the derivative of RXR_X with respect to xx and set it to zero: ddx[(24−x)⋅x]=0\frac{d}{dx}[(24 - x) \cdot x] = 0
step 4
Simplifying the derivative: ddx[24x−x2]=24−2x=0\frac{d}{dx}[24x - x^2] = 24 - 2x = 0
step 5
Solving for xx: 24−2x=0  ⟹  x=1224 - 2x = 0 \implies x = 12
step 6
Similarly, the revenue for good YY is RY=PY⋅y=(32−y)⋅yR_Y = P_Y \cdot y = (32 - y) \cdot y
step 7
To find the maximum revenue, we take the derivative of RYR_Y with respect to yy and set it to zero: ddy[(32−y)⋅y]=0\frac{d}{dy}[(32 - y) \cdot y] = 0
step 8
Simplifying the derivative: ddy[32y−y2]=32−2y=0\frac{d}{dy}[32y - y^2] = 32 - 2y = 0
step 9
Solving for yy: 32−2y=0  ⟹  y=1632 - 2y = 0 \implies y = 16
Answer
The optimal levels of production are x=12x = 12 and y=16y = 16.
Key Concept
Revenue Maximization
Explanation
To find the optimal levels of production for goods XX and YY, we maximize the revenue functions by taking the derivative with respect to each variable and setting it to zero. This gives us the values of xx and yy that maximize revenue.
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