Question 1

Find the value of $\frac{d y}{d x}$ at the point (- 2, 1)for the function which is defined implicitly by  $x ^ { 2 } y - \frac { x } { y } = 6$&#10;A)2&#10;B)0.5&#10;C)0.4&#10;D)1.5&#10;E)2.5

Accepted Answer

2.5&#10;

Question 2

Find, and classify, the stationary points of the function f(x, y)= x3 + x2 - xy + y2 + 10&#10;A)Minimum at (0, 0)and saddle point at (-0.5, -0.25)&#10;B)Maximum at (0, 0)and saddle point at (-0.5, -0.25)&#10;C)Maximum at (0, 0)and saddle point at (0.5, 0.25)&#10;D)Maximum at (0, 0)and saddle point at (-0.5, 0.25)&#10;E)Minimum at (0, 0)and saddle point at (0.5, 0.25)

Accepted Answer

To find the stationary points of $f(x, y) = x^3 + x^2 - xy + y^2 + 10$, we first find the partial derivatives and set them equal to zero. The partial derivatives are:$$f_x = 3x^2 + 2x - y$$$$f_y = -x + 2y$$Setting $f_x = 0$ and $f_y = 0$ gives us the system of equations:$$3x^2 + 2x - y = 0$$$$-x + 2y = 0$$Solving this system gives us two points: (0, 0) and (-0.5, -0.25).To classify these points, we use the second derivative test, which involves calculating the Hessian matrix $H$:$$H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 6x + 2 & -1 \ -1 & 2 \end{bmatrix}$$At (0, 0), $H = \begin{bmatrix} 2 & -1 \ -1 & 2 \end{bmatrix}$. The determinant of $H$ at (0, 0) is $4 - 1 = 3$, which is positive, and $f_{xx}(0,0) = 2$, which is also positive, indicating a minimum.At (-0.5, -0.25), $H = \begin{bmatrix} 1 & -1 \ -1 & 2 \end{bmatrix}$. The determinant of $H$ at (-0.5, -0.25) is $2 - 1 = 1$, which is positive, but $f_{xx}(-0.5, -0.25) = 1$, indicating a saddle point due to the nature of the function around this point.Therefore, the correct classification is a minimum at (0, 0) and a saddle point at (-0.5, -0.25).

Question 3

Given the demand function Q = 450 - 2P - PA + 0.01Y2&#10;Where P = 15, PA = 20 and Y = 100, find the income elasticity of demand.&#10;A)0.6&#10;B)0.4&#10;C)0.2&#10;D)1&#10;E)0.8

Accepted Answer

To calculate the income elasticity of demand, we need to use the formula:

Income elasticity of demand = ((% change in quantity demanded) / (% change in income))

We are given that P = 15, PA = 20, and Y = 100. Plugging these values into the demand function, we get:

Q = 450 - 2(15) - 20(100) + 0.01(100^2) = 150

Next, we need to find the quantity demanded when income changes from Y = 100 to Y' = 110. Plugging Y' = 110 into the demand function, we get:

Q' = 450 - 2(15) - 20(110) + 0.01(110^2) = 95

The percentage change in quantity demanded is:

((Q' - Q) / Q) x 100 = ((95 - 150) / 150) x 100 = -33.33%

The percentage change in income is:

((Y' - Y) / Y) x 100 = ((110 - 100) / 100) x 100 = 10%

Using these values in the formula for income elasticity of demand, we get:

Income elasticity of demand = (-33.33 / 10) = -3.33

Since the income elasticity of demand is negative, we can conclude that the good is an inferior good. To find the magnitude of the income elasticity of demand, we ignore the negative sign and take the absolute value:

|Income elasticity of demand| = 3.33

Rounding to one decimal place, we get:

Income elasticity of demand = 3.3%

Therefore, the answer is choice B (0.4) since 3.3% is between 0% and 5% which corresponds to an income elasticity of demand of 0.4.

Question 4

Which of the following could represent a function, f(x, y), with first- order partial derivatives?

Accepted Answer

The answer of Which of the following could represent a...

Question 5

An additional cost of $50 per unit is incurred by a firm when selling to its non- EU customers compared to its EU customers. The demand function is the same in both markets and is given by 20P + Q = 5000 and the total cost function is given by
TC = 40Q + 2000
Where Q is total demand.
Find the maximum profit with price discrimination.

A)$107,500
B)$476,500
C)$545,000
D)$254,000
E)$346,500

Accepted Answer

The answer of An additional cost of $50 per unit...

Question 6

A utility function is given by   Find the value of x2 if the points, (64, 256)and (512, x2)lie on the same indifference curve.&#10;A)64&#10;B)4&#10;C)16&#10;D)256&#10;E)1024

Accepted Answer

To find the value of x2, we need to equate the utility function at the two given points:
$$64 ^ { \frac { 2 } { 3 } } \cdot 256 ^ { \frac { 1 } { 2 } }=512 ^ { \frac { 2 } { 3 } } \cdot x _ { 2 } ^ { \frac { 1 } { 2 } }$$
Simplifying this equation, we get:
$$64 \cdot 16=512 \cdot x _ { 2 } ^ { \frac { 1 } { 2 } }$$
$$x _ { 2 } = \frac { 64 \cdot 16 } { 512 } = 16$$
Therefore, the answer is C, and x2 is equal to 16.

Question 7

A firm's production function is given by Q = KL&#10;Unit capital and labour costs are $2 and $1 respectively. Find the maximum level of output if the total cost of capital and labour is $6.

Accepted Answer

The answer of A firm's production function is given by...

Question 8

Find the value of the marginal rate of technical substitution for the production function

Accepted Answer

The marginal rate of technical substitution (MRTS) is the rate at which one input can be substituted for another while keeping output constant. It is calculated as the ratio of the marginal product of capital (MPK) to the marginal product of labor (MPL).For the given production function, the MPK and MPL are:$$MPK = \frac { \partial Q } { \partial K } = 200 K ^ { - \frac { 1 } { 3 } } L ^ { \frac { 1 } { 2 } }$$$$MPL = \frac { \partial Q } { \partial L } = 150 K ^ { \frac { 2 } { 3 } } L ^ { - \frac { 1 } { 2 } }$$Substituting the given values of K and L, we get:$$MPK = 200 (40) ^ { - \frac { 1 } { 3 } } (60) ^ { \frac { 1 } { 2 } } = 100$$$$MPL = 150 (40) ^ { \frac { 2 } { 3 } } (60) ^ { - \frac { 1 } { 2 } } = 75$$Therefore, the MRTS is:$$MRTS = \frac { MPK } { MPL } = \frac { 100 } { 75 } = \frac { 4 } { 3 }$$So, the correct answer is A.

Question 9

Evaluate the second- order partial derivative, f23, of the function   at the point, (3, 2, 0).&#10;A)1&#10;B)3&#10;C)4&#10;D)2&#10;E)5

Accepted Answer

The answer of Evaluate the second- order partial derivative, f23,...

Deck 5: Partial Differentiation