Question

In: Economics

1. The function f(x, y) = ln(x3 + 2) / (y2 + 3) (this function is...

1. The function f(x, y) = ln(x³ + 2) / (y² + 3) (this function is of a fraction format) :

	a.	has a stationary point at (1, 0)
	b.	has a stationary point at (0, 0)
	c.	has a stationary point at (0, 1)
	d.	has no stationary points

2. Which of the following functions don’t have unit elasticity at P = 6?

	a.	Demand: Q_d = 24 - 2 P
	b.	Demand: Q_d = 10/P
	c.	Demand: log Q_d = 100 - 3 log P
	d.	Supply: Q_s = 5 P

3. For the following total cost function, find marginal cost (MC), average total cost (ATC) and average variable cost (AVC) and then decide which of the following are true. TC = 0.1Q³ - 4Q² + 80Q + 150 doesn’t implies

	a.	MC = 0.3Q² - 8Q + 80
	b.	ATC = 0.1Q² - 4Q + 80
	c.	AVC = 0.1Q² - 4Q + 80
	d.	MC=AVC at Q=20

4. Solve the output production maximizing problem max Q(x, y) = −x³ − 3y² + 3x² + 24y where x and y are the necessary inputs. Find the maximum production

	a.	48
	b.	52
	c.	20
	d.	50

Expert Solution

Given , f(x,y) = [ln(x3 + 2 )]/ (y2 + 3)

The stationary point for a function of two variables lies where the partial derivatives(wr.t xand wrt y ) are zero simultaneously.

Thus df/dx = 3x2 * (y2 + 3) / [[ln(x3 + 2 )]* (y2 + 3) ] = 0

And df/dy = -[ln(x3 + 2 )]* 2y / (y2 + 3) 2 =0

= > 3x2 * (y2 + 3) = 0

And

[ln(x3 + 2 )]* 2y = 0

* Note that the denominators of both the partial derivatives are always positive

= > Either 3x2 = 0 OR (y2 + 3) = 0 or both =0

But y2 >= 0 always hence not possible

Thus 3x2 = 0 = > x = 0………………………….stationary point at (0,0)

Again from [ln(x3 + 2 )]* 2y = 0

= > either [ln(x3 + 2 )] = 0 OR 2y = 0 or both

If [ln(x3 + 2 )] = 0 = > x3 + 2= 1 = > x = -1

And also y = 0

Hence , optionb

2. the elasticity is equal to (dQ/dp)*(P/Q) where P = 6 and Q is the quantity obtained when P=6

All the Functions except c ) logQd = 100 – 3logP has a unit elasticity at P= 6

Function c) on the other hand has an elasticity of 3

Differentiating logQd = 100 – 3logP totally :-

dQd /Qd = -3*dP/P

= > (dQd/dp)*(P/Qd) = -3(Qd/P) *(P/Qd) = -3

3.

Option b is the answer

Since ATC = TC/Q = 0.1Q2 - 4Q + 80 + 150/Q

4. First order conditions are dQ/dx= 0 and dQ/dy = 0

= >-3x2 + 6x = 0 = > x (6-3x) = 0

Since x is necessary good so x not equal to zer0

Therefore, (6-3x) = 0

= > x = 2

dQ/dy = -6y+24 = 0 => y = 4

Hence max production = -23 – 3*42 + 3*22 +24*4 = -8 – 48 + 12+96 = 52 option B

Rahul Sunny answered 2 years ago

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