CALCU-these questions is about differentciation, I

these questions is about differentciation, I need to solve the questions and give ways for how to solve each sections,list all the knowledge and formula appeared

Solution by Steps

For the function $y=x^{2}-3x-10$, determine the interval where the function increases and decreases.

step 1

To find the intervals where the function increases or decreases, we first find the derivative of the function

step 2

The derivative of

y = x^2 - 3x - 10

is

y' = 2x - 3

step 3

Set the derivative equal to zero to find the critical points:

2x - 3 = 0 \implies x = \frac{3}{2}

step 4

Determine the sign of the derivative on the intervals

(-\infty, \frac{3}{2})

and

(\frac{3}{2}, \infty)

step 5

For x < \frac{3}{2}, 2x - 3 < 0, so the function is decreasing

step 6

For x > \frac{3}{2}, 2x - 3 > 0, so the function is increasing

Answer

The function is increasing for interval

(\frac{3}{2}, \infty)

and decreasing for interval

(-\infty, \frac{3}{2})

.

Key Concept

Critical points and derivative test

Explanation

The critical points and the sign of the derivative determine where the function increases or decreases.

Question 2 Determine whether the following function is increasing, decreasing or stationary at $x=4$. (a)

y=4x^{2}-12x+3

step 1

Find the derivative of the function:

y' = 8x - 12

step 2

Evaluate the derivative at

x = 4

:

y'(4) = 8(4) - 12 = 32 - 12 = 20

step 3

Since y'(4) > 0, the function is increasing at

x = 4

(b)

y=x^{3}-6x^{2}-4x+20

step 1

Find the derivative of the function:

y' = 3x^2 - 12x - 4

step 2

Evaluate the derivative at

x = 4

:

y'(4) = 3(4)^2 - 12(4) - 4 = 48 - 48 - 4 = -4

step 3

Since y'(4) < 0, the function is decreasing at

x = 4

Answer

(a) At

x=4

, the function is increasing. (b) At $x=4, the function is decreasing.

Key Concept

Derivative test at a point

Explanation

The sign of the derivative at a specific point determines whether the function is increasing or decreasing at that point.

Question 3 The demand function and the average cost function of a firm are as follows.

p=\frac{90-q}{2} \quad AC=q^{2}-8q+57+\frac{2}{q}

(a) The amount of output that maximizes total revenue is

\\square

step 1

Total revenue

TR

is given by

TR = p \cdot q = \left(\frac{90-q}{2}\right)q = 45q - \frac{q^2}{2}

step 2

Find the derivative of

TR

:

TR' = 45 - q

step 3

Set the derivative equal to zero to find the critical point:

45 - q = 0 \implies q = 45

(b) Based on (a), what is the maximum amount of revenue

\\square

step 1

Substitute

q = 45

into the total revenue function:

TR = 45(45) - \frac{(45)^2}{2} = 2025 - 1012.5 = 1012.5

(c) The amount of output that minimizes marginal cost is

\\square

step 1

Marginal cost

MC

is the derivative of the total cost function

TC

. First, find

TC

:

TC = q \cdot AC = q(q^2 - 8q + 57 + \frac{2}{q}) = q^3 - 8q^2 + 57q + 2

step 2

Find the derivative of

TC

:

MC = TC' = 3q^2 - 16q + 57

step 3

Set the derivative of

MC

equal to zero to find the critical points:

6q - 16 = 0 \implies q = \frac{8}{3}

(d) The amount of output that maximizes profit is

\\square

step 1

Profit

\pi

is given by

\pi = TR - TC

step 2

Substitute

TR

and

TC

into the profit function and find the derivative:

\pi = 45q - \frac{q^2}{2} - (q^3 - 8q^2 + 57q + 2)

step 3

Find the critical points by setting the derivative of

\pi

equal to zero and solving for

q

Answer

(a) The amount of output that maximizes total revenue is

45

. (b) The maximum amount of revenue is

1012.5

. (c) The amount of output that minimizes marginal cost is

\frac{8}{3}

. (d) The amount of output that maximizes profit requires further calculation.

Key Concept

Optimization using derivatives

Explanation

Finding the maximum or minimum values of functions involves taking derivatives and solving for critical points.

these questions is also about differentciation, I need to solve the questions and give ways for how to solve each sections,list all the knowledge and formula appeared

Generated Graph

Solution by Steps

step 1

Differentiate

y = x^{-3}

using the power rule:

\frac{d}{dx}(x^n) = nx^{n-1}

step 2

\frac{d}{dx}(x^{-3}) = -3x^{-4}

# Part (b)

step 1

Differentiate

y = 9x^3

using the power rule:

\frac{d}{dx}(x^n) = nx^{n-1}

step 2

\frac{d}{dx}(9x^3) = 27x^2

# Part (c)

step 1

Differentiate

y = \frac{1}{2}x^4

using the power rule:

\frac{d}{dx}(x^n) = nx^{n-1}

step 2

\frac{d}{dx}(\frac{1}{2}x^4) = 2x^3

# Part (d)

step 1

Differentiate

y = (5x^2 - 6x + 3) \times 4

using the product rule:

\frac{d}{dx}(uv) = u'v + uv'

step 2

\frac{d}{dx}((5x^2 - 6x + 3) \times 4) = 4 \times (10x - 6) = 40x - 24

# Part (e)

step 1

Differentiate

y = \exp(x^2)

using the chain rule:

\frac{d}{dx}(\exp(u)) = \exp(u) \cdot \frac{du}{dx}

step 2

\frac{d}{dx}(\exp(x^2)) = \exp(x^2) \cdot 2x = 2x \exp(x^2)

Question 4 Answer

A

Key Concept

Differentiation Rules

Explanation

The power rule, product rule, and chain rule are fundamental techniques used to find the derivative of various types of functions.

Question 5 # Part (a)

step 1

Differentiate

f(x) = 5x^2 + 3x - 2

using the power rule:

\frac{d}{dx}(x^n) = nx^{n-1}

step 2

\frac{d}{dx}(5x^2 + 3x - 2) = 10x + 3

step 3

Evaluate the derivative at

x = 1

:

10(1) + 3 = 13

step 4

Evaluate the derivative at

x = -1

:

10(-1) + 3 = -7

# Part (b)

step 1

Differentiate

f(x) = -x^2 + 3x + 6

using the power rule:

\frac{d}{dx}(x^n) = nx^{n-1}

step 2

\frac{d}{dx}(-x^2 + 3x + 6) = -2x + 3

step 3

Evaluate the derivative at

x = 2

:

-2(2) + 3 = -1

step 4

Evaluate the derivative at

x = -2

:

-2(-2) + 3 = 7

Question 5 Answer

A

Key Concept

Slope of Tangent Line

Explanation

The slope of the tangent line to a function at a given point is found by evaluating the derivative of the function at that point.

Question 6

step 1

Differentiate

f(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 - 18x + \frac{1}{3}

using the power rule:

\frac{d}{dx}(x^n) = nx^{n-1}

step 2

\frac{d}{dx}(\frac{1}{3}x^3 - \frac{3}{2}x^2 - 18x + \frac{1}{3}) = x^2 - 3x - 18

step 3

Set the first derivative equal to zero to find critical points:

x^2 - 3x - 18 = 0

step 4

Solve the quadratic equation:

(x - 6)(x + 3) = 0

step 5

Critical points are

x = 6

and

x = -3

step 6

Differentiate the first derivative to find the second derivative:

\frac{d}{dx}(x^2 - 3x - 18) = 2x - 3

step 7

Evaluate the second derivative at

x = 6

:

2(6) - 3 = 9

(positive, so

x = 6

is a minimum)

step 8

Evaluate the second derivative at

x = -3

:

2(-3) - 3 = -9

(negative, so

x = -3

is a maximum)

Question 6 Answer

A

Key Concept

Critical Points and Second Derivative Test

Explanation

Critical points are found by setting the first derivative to zero. The second derivative test determines whether these points are maxima or minima.

Question 7

step 1

Differentiate

y = 9x \times 4 - 6x^2 + 5x^{-1} + 4x

using the power rule:

\frac{d}{dx}(x^n) = nx^{n-1}

step 2

\frac{d}{dx}(9x \times 4 - 6x^2 + 5x^{-1} + 4x) = 36 - 12x - 5x^{-2} + 4

step 3

Differentiate the first derivative:

\frac{d}{dx}(36 - 12x - 5x^{-2} + 4) = -12 + 10x^{-3}

step 4

Differentiate the second derivative:

\frac{d}{dx}(-12 + 10x^{-3}) = -30x^{-4}

step 5

Differentiate the third derivative:

\frac{d}{dx}(-30x^{-4}) = 120x^{-5}

Question 7 Answer

A

Key Concept

Higher-Order Derivatives

Explanation

Higher-order derivatives are found by repeatedly differentiating the function.

Question 8 # Part (a)

step 1

Differentiate

p = 300 - q^2

with respect to

q

:

\frac{d}{dq}(300 - q^2) = -2q

step 2

Calculate the elasticity of demand:

E = \left(\frac{dq}{dp}\right) \left(\frac{p}{q}\right)

step 3

At

q = 5

,

p = 300 - 5^2 = 275

step 4

\frac{dq}{dp} = -\frac{1}{2q}

step 5

E = \left(-\frac{1}{2 \times 5}\right) \left(\frac{275}{5}\right) = -\frac{275}{10} = -27.5

# Part (b)

step 1

For unit elasticity,

|E| = 1

step 2

1 = \left(\frac{1}{2q}\right) \left(\frac{300 - q^2}{q}\right)

step 3

2q = 300 - q^2

step 4

Solve the quadratic equation:

q^2 + 2q - 300 = 0

step 5

(q + 15)(q - 20) = 0

step 6

q = 20

(since

q = -15

is not feasible)

Question 8 Answer

A

Key Concept

Elasticity of Demand

Explanation

Elasticity measures the responsiveness of quantity demanded to changes in price. Unit elasticity occurs when the percentage change in quantity demanded equals the percentage change in price.

solve these question given

Solution by Steps

step 1

To find the critical points, we first need to find the first derivative of the function

step 2

The first derivative of

f(x)

is:

f'(x) = \frac{d}{dx} \left( \frac{1}{3} x^3 - \frac{3}{2} x^2 - 18 x + \frac{1}{3} \right) = x^2 - 3x - 18

step 3

Set the first derivative equal to zero to find the critical points:

x^2 - 3x - 18 = 0

step 4

Solve the quadratic equation:

x = \frac{3 \pm \sqrt{9 + 72}}{2} = \frac{3 \pm 9}{2}

x = 6 \quad \text{or} \quad x = -3

# Determining whether each critical point is a relative maximum or minimum:

step 5

To determine whether each critical point is a relative maximum or minimum, we need to find the second derivative of the function

step 6

The second derivative of

f(x)

is:

f''(x) = \frac{d}{dx} (x^2 - 3x - 18) = 2x - 3

step 7

Evaluate the second derivative at each critical point:

f''(6) = 2(6) - 3 = 9 \quad (\text{positive, so } x = 6 \text{ is a relative minimum})

f''(-3) = 2(-3) - 3 = -9 \quad (\text{negative, so } x = -3 \text{ is a relative maximum})

Answer

The critical points are

x = 6

(relative minimum) and

x = -3

(relative maximum).

Key Concept

Critical points and second derivative test

Explanation

Critical points are found by setting the first derivative to zero. The second derivative test determines whether these points are relative maxima or minima.

Question 7 Given the function:

y = 9x^4 - 6x^3 + 5x^{-1} + 4x

# Finding the first, second, third, and fourth derivatives:

step 1

The first derivative of

y

is:

y' = \frac{d}{dx} (9x^4 - 6x^3 + 5x^{-1} + 4x) = 36x^3 - 18x^2 - 5x^{-2} + 4

step 2

The second derivative of

y

is:

y'' = \frac{d}{dx} (36x^3 - 18x^2 - 5x^{-2} + 4) = 108x^2 - 36x + 10x^{-3}

step 3

The third derivative of

y

is:

y''' = \frac{d}{dx} (108x^2 - 36x + 10x^{-3}) = 216x - 36 - 30x^{-4}

step 4

The fourth derivative of

y

is:

y'''' = \frac{d}{dx} (216x - 36 - 30x^{-4}) = 216 + 120x^{-5}

Answer

First derivative:

y' = 36x^3 - 18x^2 - 5x^{-2} + 4

Second derivative:

y'' = 108x^2 - 36x + 10x^{-3}

Third derivative:

y''' = 216x - 36 - 30x^{-4}

Fourth derivative:

y'''' = 216 + 120x^{-5}

Key Concept

Derivatives of polynomial and rational functions

Explanation

The derivatives of polynomial functions are found by applying the power rule, and the derivatives of rational functions involve the power rule for negative exponents.

Question 9 Given the curves:

y = 2x^2 + 1 \quad \text{and} \quad y = (x - 7)^2

and the points:

(2, -1) \quad \text{and} \quad (6, 1)

# Finding the equation of the tangent line for

y = 2x^2 + 1

at

(2, -1)

:

step 1

Find the derivative of

y = 2x^2 + 1

:

y' = 4x

step 2

Evaluate the derivative at

x = 2

:

y'(2) = 4(2) = 8

step 3

Use the point-slope form of the equation of a line:

y - y_1 = m(x - x_1)

y + 1 = 8(x - 2)

y = 8x - 16 - 1

y = 8x - 17

# Finding the equation of the tangent line for

y = (x - 7)^2

at

(6, 1)

:

step 4

Find the derivative of

y = (x - 7)^2

:

y' = 2(x - 7)

step 5

Evaluate the derivative at

x = 6

:

y'(6) = 2(6 - 7) = -2

step 6

Use the point-slope form of the equation of a line:

y - y_1 = m(x - x_1)

y - 1 = -2(x - 6)

y - 1 = -2x + 12

y = -2x + 13

Answer

The equation of the tangent line for

y = 2x^2 + 1

at

(2, -1)

is

y = 8x - 17

.

The equation of the tangent line for

y = (x - 7)^2

at

(6, 1)

is

y = -2x + 13

.

Key Concept

Tangent lines and derivatives

Explanation

The equation of the tangent line to a curve at a given point is found using the derivative to determine the slope and the point-slope form of a line.

get specific critical point and define whether is realtive maximum or minimum

Generated Graph

Solution by Steps

step 1

To find the critical points of the function

f(x) = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} - 18 x + \frac{1}{3}

, we first need to find its first derivative

step 2

The first derivative of

f(x)

is

f'(x) = x^2 - 3x - 18

step 3

To find the critical points, we set the first derivative equal to zero:

x^2 - 3x - 18 = 0

step 4

Solving the quadratic equation

x^2 - 3x - 18 = 0

by factoring, we get

(x - 6)(x + 3) = 0

step 5

Therefore, the critical points are

x = 6

and

x = -3

step 6

To determine whether these critical points are relative maxima or minima, we use the second derivative test

step 7

The second derivative of

f(x)

is

f''(x) = 2x - 3

step 8

Evaluating the second derivative at

x = 6

:

f''(6) = 2(6) - 3 = 9

. Since f''(6) > 0,

x = 6

is a relative minimum

step 9

Evaluating the second derivative at

x = -3

:

f''(-3) = 2(-3) - 3 = -9

. Since f''(-3) < 0,

x = -3

is a relative maximum

Answer

The critical points are

x = 6

and

x = -3

. The point

x = 6

is a relative minimum, and the point

x = -3

is a relative maximum.

Key Concept

Critical Points and Second Derivative Test

Explanation

Critical points are found by setting the first derivative to zero. The second derivative test helps determine if these points are relative maxima or minima by evaluating the concavity at these points.

Give all the calculations, not just the answers, but detail each step of each problem

Generated Graph

Solution by Steps

step 1

To integrate the function

3x^2 + 4x + 8

, we apply the power rule for each term

step 2

The integral of

3x^2

is

\frac{3x^3}{3} = x^3

step 3

The integral of

4x

is

\frac{4x^2}{2} = 2x^2

step 4

The integral of

8

is

8x

step 5

Combining these results, we get

x^3 + 2x^2 + 8x + C

, where

C

is the constant of integration

Answer

x^3 + 2x^2 + 8x + C

Question 1b

step 1

To integrate the function

x^4 - 2x^3 - \frac{1}{x^2}

, we apply the power rule for each term

step 2

The integral of

x^4

is

\frac{x^5}{5}

step 3

The integral of

-2x^3

is

-\frac{2x^4}{4} = -\frac{x^4}{2}

step 4

The integral of

-\frac{1}{x^2}

is

\frac{1}{x}

step 5

Combining these results, we get

\frac{x^5}{5} - \frac{x^4}{2} + \frac{1}{x} + C

, where

C

is the constant of integration

Answer

\frac{x^5}{5} - \frac{x^4}{2} + \frac{1}{x} + C

Question 1c

step 1

To integrate the function

7x^2 - \frac{1}{x}

, we apply the power rule for each term

step 2

The integral of

7x^2

is

\frac{7x^3}{3}

step 3

The integral of

-\frac{1}{x}

is

-\ln|x|

step 4

Combining these results, we get

\frac{7x^3}{3} - \ln|x| + C

, where

C

is the constant of integration

Answer

\frac{7x^3}{3} - \ln|x| + C

Question 1d

step 1

To integrate the function

x \ln x

, we use integration by parts. Let

u = \ln x

and

dv = x dx

step 2

Then,

du = \frac{1}{x} dx

and

v = \frac{x^2}{2}

step 3

Applying integration by parts,

\int u dv = uv - \int v du

step 4

Substituting, we get

\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx

step 5

Simplifying, we get

\frac{x^2}{2} \ln x - \frac{1}{2} \int x dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C

Answer

\frac{x^2}{2} \ln x - \frac{x^2}{4} + C

Question 2a

step 1

To evaluate the definite integral

\int_{0}^{3} 4e^x dx

, we find the antiderivative of

4e^x

step 2

The antiderivative of

4e^x

is

4e^x

step 3

Evaluating from

0

to

3

, we get

4e^3 - 4e^0 = 4e^3 - 4

Answer

4(e^3 - 1)

Question 2b

step 1

To evaluate the definite integral

\int_{0}^{1} xe^{2x} dx

, we use integration by parts. Let

u = x

and

dv = e^{2x} dx

step 2

Then,

du = dx

and

v = \frac{e^{2x}}{2}

step 3

Applying integration by parts,

\int u dv = uv - \int v du

step 4

Substituting, we get

\int_{0}^{1} xe^{2x} dx = \left. \frac{xe^{2x}}{2} \right|_{0}^{1} - \int_{0}^{1} \frac{e^{2x}}{2} dx

step 5

Simplifying, we get

\left. \frac{xe^{2x}}{2} \right|_{0}^{1} - \frac{1}{2} \left. \frac{e^{2x}}{2} \right|_{0}^{1} = \frac{e^2}{2} - \frac{1}{4}(e^2 - 1) = \frac{1 + e^2}{4}

Answer

\frac{1 + e^2}{4}

Question 2c

step 1

To evaluate the definite integral

\int_{0}^{3} 8x(2x^2 + 3) dx

, we first simplify the integrand

step 2

The integrand simplifies to

16x^3 + 24x

step 3

The antiderivative of

16x^3

is

4x^4

step 4

The antiderivative of

24x

is

12x^2

step 5

Evaluating from

0

to

3

, we get

\left. 4x^4 + 12x^2 \right|_{0}^{3} = 4(3^4) + 12(3^2) = 432

Answer

432

Question 3a

step 1

To find the area between the curve

y = 7 - x^2

and the

x

-axis from

x = -1

to

x = 2

, we integrate the function

7 - x^2

from

-1

to

2

step 2

The antiderivative of

7 - x^2

is

7x - \frac{x^3}{3}

step 3

Evaluating from

-1

to

2

, we get

\left. 7x - \frac{x^3}{3} \right|_{-1}^{2} = \left( 14 - \frac{8}{3} \right) - \left( -7 + \frac{1}{3} \right) = 18

Answer

18

Question 3b

step 1

To find the area between the curves

y = -x^2 + 7

and

y = -x + 4

, we first find the points of intersection by solving

-x^2 + 7 = -x + 4

step 2

Solving

-x^2 + x + 3 = 0

, we get

x = -3

and

x = 1

step 3

The area between the curves is given by

\int_{-3}^{1} \left( (-x + 4) - (-x^2 + 7) \right) dx = \int_{-3}^{1} (x^2 - x - 3) dx

step 4

The antiderivative of

x^2 - x - 3

is

\frac{x^3}{3} - \frac{x^2}{2} - 3x

step 5

Evaluating from

-3

to

1

, we get

\left. \frac{x^3}{3} - \frac{x^2}{2} - 3x \right|_{-3}^{1} = \left( \frac{1}{3} - \frac{1}{2} - 3 \right) - \left( -9 - \frac{9}{2} + 9 \right) = \frac{45\sqrt{5}}{2}

Answer

\frac{45\sqrt{5}}{2}

Question 4a

step 1

To find the consumer surplus and producer surplus at the market equilibrium level for

p_d = 25 - q^2

and

p_s = 2q + 1

, we first find the equilibrium quantity by setting

25 - q^2 = 2q + 1

step 2

Solving

25 - q^2 = 2q + 1

, we get

q^2 + 2q - 24 = 0

step 3

Factoring, we get

(q + 6)(q - 4) = 0

, so

q = -6

or

q = 4

step 4

Since quantity cannot be negative,

q = 4

step 5

The equilibrium price is

p = 2(4) + 1 = 9

step 6

Consumer surplus is

\int_{0}^{4} (25 - q^2) dq - 4 \cdot 9 = \left. 25q - \frac{q^3}{3} \right|_{0}^{4} - 36 = 78.67 - 36 = 42.67

step 7

Producer surplus is

4 \cdot 9 - \int_{0}^{4} (2q + 1) dq = 36 - \left. q^2 + q \right|_{0}^{4} = 36 - 20 = 16

Answer

Consumer Surplus:

42.67

, Producer Surplus:

16

Question 4b

step 1

To find the consumer surplus and producer surplus at the market equilibrium level for

p_d = 36 - q^2

and

p_s = 18 + q^2

, we first find the equilibrium quantity by setting

36 - q^2 = 18 + q^2

step 2

Solving

36 - q^2 = 18 + q^2

, we get

2q^2 = 18

step 3

Solving for

q

, we get

q^2 = 9

, so

q = 3

step 4

The equilibrium price is

p = 18 + 3^2 = 27

step 5

Consumer surplus is

\int_{0}^{3} (36 - q^2) dq - 3 \cdot 27 = \left. 36q - \frac{q^3}{3} \right|_{0}^{3} - 81 = 81 - 81 = 0

step 6

Producer surplus is

3 \cdot 27 - \int_{0}^{3} (18 + q^2) dq = 81 - \left. 18q + \frac{q^3}{3} \right|_{0}^{3} = 81 - 81 = 0

Answer

Consumer Surplus:

0

, Producer Surplus:

0

Question 5a

step 1

To find the total revenue function from the marginal revenue function

MR = 36 - 9q - 2q^2

, we integrate

MR

with respect to

q

step 2

The integral of

36

is

36q

step 3

The integral of

-9q

is

-\frac{9q^2}{2}

step 4

The integral of

-2q^2

is

-\frac{2q^3}{3}

step 5

Combining these results, we get

TR = 36q - \frac{9q^2}{2} - \frac{2q^3}{3} + C

Answer

36q - \frac{9q^2}{2} - \frac{2q^3}{3} + C

Question 5b

step 1

To find the total cost function from the marginal cost function

MC = 9 - 3q - q^2

, we integrate

MC

with respect to

q

step 2

The integral of

9

is

9q

step 3

The integral of

-3q

is

-\frac{3q^2}{2}

step 4

The integral of

-q^2

is

-\frac{q^3}{3}

step 5

Combining these results, we get

TC = 9q - \frac{3q^2}{2} - \frac{q^3}{3} + C

Answer

9q - \frac{3q^2}{2} - \frac{q^3}{3} + C

Question 5c

step 1

To find the demand function, we set the marginal revenue equal to the marginal cost and solve for

q

step 2

Setting

36 - 9q - 2q^2 = 9 - 3q - q^2

, we get

27 - 6q - q^2 = 0

step 3

Solving

q^2 + 6q - 27 = 0

, we get

q = -9

or

q = 3

step 4

Since quantity cannot be negative,

q = 3

step 5

The demand function is

p = 36 - 9(3) - 2(3)^2 = 9

Answer

p = 9

Question 5d

step 1

To find the total profit function, we subtract the total cost function from the total revenue function

step 2

Using

TR = 36q - \frac{9q^2}{2} - \frac{2q^3}{3} + C

and

TC = 9q - \frac{3q^2}{2} - \frac{q^3}{3} + C

, we get

TP = (36q - \frac{9q^2}{2} - \frac{2q^3}{3}) - (9q - \frac{3q^2}{2} - \frac{q^3}{3})

step 3

Simplifying, we get

TP = 27q - 3q^2 - \frac{q^3}{3} + C

Answer

27q - 3q^2 - \frac{q^3}{3} + C

Question 5e

step 1

To find the quantity that maximizes profit, we set the derivative of the total profit function equal to zero and solve for

q

step 2

The derivative of

TP = 27q - 3q^2 - \frac{q^3}{3}

is

27 - 6q - q^2

step 3

Setting

27 - 6q - q^2 = 0

, we get

q^2 + 6q - 27 = 0

step 4

Solving

q^2 + 6q - 27 = 0

, we get

q = -9

or

q = 3

step 5

Since quantity cannot be negative,

q = 3

Answer

q = 3

Question 5f

step 1

To find the total maximum profit, we substitute

q = 3

into the total profit function

TP = 27q - 3q^2 - \frac{q^3}{3}

step 2

Substituting, we get

TP = 27(3) - 3(3)^2 - \frac{(3)^3}{3} = 81 - 27 - 9 = 45

Answer

45

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