An increase in the real wage rate ________ the quantity of labor demanded, ______ the quantity of labor supplied, and when the l
1 answer:
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Answer:
<u>Part (a):</u>
Make a table showing the value of the marginal product for each screen from the first through the fifth:
<u>Solution: </u>
The answer is attached.
<u>Part (b):</u>
How many screens will be built if the real interest rate is 5.5 percent?
<u>Answer:</u> 3 screens
<u>Part (c): </u>
How many screens will be built if the real interest rate is 7.5 percent?
<u>Answer:</u> 1 screen
<u>Part (d):</u>
How many screens will be built if the real interest rate is 10 percent?
<u>Answer:</u> 0 screens
<u>Part (e): </u>
If the real interest rate is 5.5 percent, how far would construction costs have to fall before the builder would be willing to build a five-screen complex?
<u>Answer:</u> $727,272.73(approx.)
Explanation:
Part (a):
Make a table showing the value of the marginal product for each screen from the first through the fifth:
Solution:
The solution is attached with working.
<u>Part (b):</u>
<u>How many screens will be built if the real interest rate is 5.5 percent?</u>
<u>Solution:</u>
3 screens
The interest cost of each screen = 5.5% x $1,000,000 = $55,000.
There are no other costs mentioned. The value of marginal product exceeds $55,000 for 3 screens.
Therefore, 3 screens should be built.
<u>Part (c): </u>
<u>How many screens will be built if the real interest rate is 7.5 percent?</u>
<u>Solution:</u>
1 screen
The value of the marginal product exceeds the interest cost (7.5% of $1,000,000, or $75,000) for only the first screen.
Thus, <u>one</u> screen will be built.
<u>Part (d):</u>
<u>How many screens will be built if the real interest rate is 10 percent?</u>
<u>Solution:</u>
0 screens
At 10% interest, the interest cost of a screen is $100,000, more than the value of the marginal product of even the first screen.
<u> </u>Thus, no screens will be built.
Part (e):
<u>If the real interest rate is 5.5 percent, how far would construction costs have to fall before the builder would be willing to build a five-screen complex?</u>
<u>Solution:</u>
The value of the marginal product of the fifth screen is $40,000. At an interest rate of 5.5%, building five screens is profitable only if 5.5% times the per-screen construction cost is no greater than $40,000.
<u>Financial cost per screen = real interest rate x construction cost of per screen </u>
$40, 000 = 5.5% x construction cost per screen Construction cost per screen = $40,000 ÷ 5.5%
= $727,272.73(approx.)
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Answer:
(A) p = -0.002x +6; 0 ≤ x ≤ 3000
(B) R(x) = x(6 -0.002x); 0 ≤ x ≤ 3000
(C) C(x) = 1.44x +1903
(D) (550, 2695), (1730, 4394.20)
(E) P(x) = -0.002x^2 +4.56x -1903
(F) increasing at $2.16 per hamburger
Explanation:
(A) The two-point form of the equation for a line can be used.
y = (y2 -y1)/(x2 -x1)(x -x1) +y1
The two points we have are ...
(x, p) = {(1750, 2.50), (1450, 3.10)}
so the equation is ...
p = (3.10 -2.50)/(1450 -1750)/(x -1750) +2.50
p = 0.6/-300(x -1750) +2.50
p = -0.002x +6
The domain of this function is where x and p are greater than 0. That will be for ...
0 ≤ x ≤ 3000
__
(B) Revenue is the product of burgers sold (x) and their price (p).
R(x) = xp
R(x) = x(6 -0.002x)
The domain of R(x) is 0 ≤ x ≤ 3000. This is the same as the domain of p(x).
___
(C) The cost function is the sum of fixed costs and variable costs:
C(x) = 1.44x +1903
__
(D) See the attachment for a graph of cost and revenue. The break-even points are (x, revenue) = (550, 2695), (1730, 4394.20).
__
(E) Profit is the difference between revenue and cost.
P(x) = R(x) - C(x) = x(6 -0.002x) -(1.44x +1903)
P(x) = -0.002x^2 +4.56x -1903
__
(F) The marginal profit is the derivative of the profit function:
P'(x) = -0.004x +4.56
P'(600) = -0.004(600) +4.56 = -2.40 +4.56 = 2.16
At a production level of 600, the profit is increasing at a rate of $2.16 per hamburger.
