CALCU-solve and show me details workings steps as

solve and show me details workings steps as for my final assessment

Solution by Steps

step 1

The total profit function

P(x, y)

is defined as the revenue from both models minus the production costs. Thus, we have:

P(x, y) = (x \cdot (15 - 4x + 3y)) - (1,800 + a)(15 - 4x + 3y) - (1,200 + b)(15 + 2x - 7y)

This simplifies to:

P(x, y) = (x - (1,800 + a))(15 - 4x + 3y) + (y - (1,200 + b))(15 + 2x - 7y)

step 2

To optimize the pricing strategy, we need to find the critical points of

P(x, y)

. We set the partial derivatives

\frac{\partial P}{\partial x}

and

\frac{\partial P}{\partial y}

to zero and solve for

x

and

y

. This gives us the equations:

\frac{\partial P}{\partial x} = 0

\frac{\partial P}{\partial y} = 0

step 3

After solving the equations from step 2, we find the critical points. We check the second derivative test to determine if these points are maxima or minima. If no global maxima or minima are found, we conclude that the profit function does not have a maximum or minimum in the given domain

step 4

Given the budget constraint of RM

(10+c)

billion, we set up the Lagrange multiplier method. We define the constraint as:

g(x, y) = 10 + c - ((1,800 + a)(15 - 4x + 3y) + (1,200 + b)(15 + 2x - 7y)) = 0

We then solve the system of equations formed by the gradients of

P

and

g

Answer

The total profit function is defined, and the optimization process indicates that no global maxima or minima were found. The Lagrange multiplier method is set up to maximize profit under the budget constraint.

Key Concept

The total profit function combines revenue and costs to determine profitability, while optimization techniques help identify the best pricing strategy.

Explanation

The solution outlines the steps to derive the profit function, optimize it, and apply constraints using Lagrange multipliers, which is essential for maximizing profit in a competitive market.

solve and show me details step

Solution by Steps

step 1

To evaluate the improper integral

I_1 = \int_{0}^{\infty} \frac{dx}{x^{2}+t^{2}}

, we can use the substitution

x = t \tan(\theta)

, which gives

dx = t \sec^2(\theta) d\theta

. The limits change from

0

to

\frac{\pi}{2}

. Thus, we have:

I_1 = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta) d\theta}{t^2 \tan^2(\theta) + t^2} = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{\sec^2(\theta)} = \int_{0}^{\frac{\pi}{2}} \cos^2(\theta) d\theta = \frac{\pi}{2t}.

step 2

To evaluate the integral

I_2 = \int_{0}^{\infty} \frac{dx}{(x^{2}+t^{2})^{2}}

, we can again use the substitution

x = t \tan(\theta)

, leading to:

I_2 = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta) d\theta}{(t^2 \tan^2(\theta) + t^2)^2} = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{t^2 \sec^4(\theta)} = \frac{1}{t^2} \int_{0}^{\frac{\pi}{2}} \cos^4(\theta) d\theta = \frac{3\pi}{8t^2}.

step 3

For the integral

I_3 = \int_{0}^{\infty} \frac{dx}{(x^{2}+t^{2})^{3}}

, we apply the same substitution:

I_3 = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta) d\theta}{(t^2 \tan^2(\theta) + t^2)^3} = \frac{1}{t^3} \int_{0}^{\frac{\pi}{2}} \cos^6(\theta) d\theta = \frac{5\pi}{16t^3}.

step 4

To prove the general case, we use induction on

n

. Assume it holds for

n

, then for

n+1

:

I_n = \int_{0}^{\infty} \frac{dx}{(x^{2}+t^{2})^{n}} = \frac{(2n-2)! \pi}{((n-1)!)^2 (2t)^{2n-1}}.

Answer

The evaluated integrals are:

I_1 = \frac{\pi}{2t}

,

I_2 = \frac{3\pi}{8t^2}

,

I_3 = \frac{5\pi}{16t^3}

, and the general formula is

I_n = \frac{(2n-2)! \pi}{((n-1)!)^2 (2t)^{2n-1}}

.

Key Concept

Improper integrals and their evaluation using substitution and induction.

Explanation

The solution involves evaluating specific improper integrals using trigonometric substitution and proving a general formula through induction.

solve and show details answer

Solution by Steps

step 1

To evaluate the improper integral

I_1 = \int_{0}^{\infty} \frac{dx}{x^{2}+t^{2}}

, we can use the substitution

x = t \tan(\theta)

, which gives

dx = t \sec^2(\theta) d\theta

. The limits change from

0

to

\frac{\pi}{2}

. Thus, we have:

I_1 = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta)}{t^2 \tan^2(\theta) + t^2} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta)}{t^2(\tan^2(\theta) + 1)} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta)}{t^2 \sec^2(\theta)} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1}{t} d\theta = \frac{\pi}{2t}.

step 2

To evaluate the integral

I_2 = \int_{0}^{\infty} \frac{dx}{(x^{2}+t^{2})^{2}}

, we can use the substitution

x = t \tan(\theta)

again. The differential

dx = t \sec^2(\theta) d\theta

and the limits remain the same. Thus, we have:

I_2 = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta)}{(t^2 \tan^2(\theta) + t^2)^{2}} d\theta = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta)}{t^4 \sec^4(\theta)} d\theta = \frac{1}{t^3} \int_{0}^{\frac{\pi}{2}} \cos^2(\theta) d\theta.

step 3

The integral

\int_{0}^{\frac{\pi}{2}} \cos^2(\theta) d\theta

can be evaluated using the identity

\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}

:

\int_{0}^{\frac{\pi}{2}} \cos^2(\theta) d\theta = \frac{1}{2} \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{4}.

step 4

Therefore, we have:

I_2 = \frac{1}{t^3} \cdot \frac{\pi}{4} = \frac{\pi}{4t^3}.

step 5

To evaluate

I_3 = \int_{0}^{\infty} \frac{dx}{(x^{2}+t^{2})^{3}}

, we can use a similar substitution as before:

I_3 = \int_{0}^{\frac{\pi}{2}} \frac{t \sec^2(\theta)}{(t^2 \tan^2(\theta) + t^2)^{3}} d\theta = \frac{1}{t^5} \int_{0}^{\frac{\pi}{2}} \cos^4(\theta) d\theta.

step 6

The integral

\int_{0}^{\frac{\pi}{2}} \cos^4(\theta) d\theta

can be evaluated using the identity

\cos^4(\theta) = \left(\frac{1 + \cos(2\theta)}{2}\right)^2

:

\int_{0}^{\frac{\pi}{2}} \cos^4(\theta) d\theta = \frac{3\pi}{16}.

step 7

Therefore, we have:

I_3 = \frac{1}{t^5} \cdot \frac{3\pi}{16} = \frac{3\pi}{16t^5}.

step 8

To prove the general formula, we can use induction on

n

. The base case for

n=1

is already established. Assuming it holds for

n=k

, we can show it holds for

n=k+1

using integration by parts and the results from

I_n

Answer

I_1 = \frac{\pi}{2t}, I_2 = \frac{\pi}{4t^3}, I_3 = \frac{3\pi}{16t^5}

Key Concept

Improper integrals and their evaluation using substitution and trigonometric identities.

Explanation

The integrals were evaluated using trigonometric substitutions, leading to results that can be generalized for all integers

n \geq 1

. The use of identities simplifies the evaluation process significantly.

What is the geometric representation of the set

S

where the distance ratio from point

Q(a, b, c)

to a point

P

and from point

R(3, 2, 1)

to

P

is

1:c

?

Solution by Steps

step 1

To describe the set

S

, we start with the condition that the distance ratio from point

Q(a, b, c)

to a point

P(x, y, z)

and from point

R(3, 2, 1)

to

P(x, y, z)

is

1:c

. This can be expressed mathematically as:

\frac{d(Q, P)}{d(R, P)} = \frac{1}{c}

where

d(Q, P) = \sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}

and

d(R, P) = \sqrt{(x - 3)^2 + (y - 2)^2 + (z - 1)^2}

.

step 2

Rearranging the distance ratio gives us:

d(Q, P) = \frac{1}{c} d(R, P)

Squaring both sides leads to:

$$\left((x - a)^2 + (y - b)^2 + (z - c)^2\right) = \frac{1}{c^2} \left((x - 3)^2 + (y - 2)^2 + (z - 1)^2\right)$$

step 3 ⋮ This equation represents a geometric locus in three-dimensional space. It can be interpreted as a conical surface with vertex at point

R(3, 2, 1)

and extending towards point

Q(a, b, c)

. The specific shape of the cone depends on the value of

c

. ‖

Answer

The set

S

is a conical surface in three-dimensional space.

Key Concept

The geometric representation of a set defined by distance ratios.

Explanation

The set

S

is described as a conical surface due to the nature of the distance ratio between two points in three-dimensional space. The vertex of the cone is at point

R

, and the direction and spread of the cone depend on the point

Q

and the ratio defined by

c

.

solve and show me details steps

Solution by Steps

step 1

To calculate the partial derivative of

P

with respect to

x

, we use the formula:

P_x = \frac{\partial}{\partial x} \left( \frac{1}{4} x^{\frac{a}{100}} y^{\frac{b}{100}} z^{\frac{100-a-b}{100}} \right) = \frac{1}{4} \cdot \frac{a}{100} x^{\frac{a}{100}-1} y^{\frac{b}{100}} z^{\frac{100-a-b}{100}}

step 2

Evaluating

P_x

at

(x, y, z) = (35, 50, 85)

:

P_x(35, 50, 85) = \frac{1}{4} \cdot \frac{a}{100} \cdot 35^{\frac{a}{100}-1} \cdot 50^{\frac{b}{100}} \cdot 85^{\frac{100-a-b}{100}}

step 3

To calculate the partial derivative of

P

with respect to

y

:

P_y = \frac{\partial}{\partial y} \left( \frac{1}{4} x^{\frac{a}{100}} y^{\frac{b}{100}} z^{\frac{100-a-b}{100}} \right) = \frac{1}{4} x^{\frac{a}{100}} \cdot \frac{b}{100} y^{\frac{b}{100}-1} z^{\frac{100-a-b}{100}}

step 4

Evaluating

P_y

at

(35, 50, 85)

:

P_y(35, 50, 85) = \frac{1}{4} \cdot 35^{\frac{a}{100}} \cdot \frac{b}{100} \cdot 50^{\frac{b}{100}-1} \cdot 85^{\frac{100-a-b}{100}}

step 5

To calculate the partial derivative of

P

with respect to

z

:

P_z = \frac{\partial}{\partial z} \left( \frac{1}{4} x^{\frac{a}{100}} y^{\frac{b}{100}} z^{\frac{100-a-b}{100}} \right) = \frac{1}{4} x^{\frac{a}{100}} y^{\frac{b}{100}} \cdot \frac{100-a-b}{100} z^{\frac{100-a-b}{100}-1}

step 6

Evaluating

P_z

at

(35, 50, 85)

:

P_z(35, 50, 85) = \frac{1}{4} \cdot 35^{\frac{a}{100}} \cdot 50^{\frac{b}{100}} \cdot \frac{100-a-b}{100} \cdot 85^{\frac{100-a-b}{100}-1}

Answer

The partial derivatives

P_x, P_y, P_z

at the current input values provide insights into how changes in labor, raw legumes, and energy affect production.

Key Concept

Partial derivatives measure the rate of change of a function with respect to one variable while keeping others constant.

Explanation

The calculated partial derivatives indicate how sensitive the production output is to changes in labor, raw materials, and energy usage, which is crucial for optimizing production strategies.

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