Group work 2/17 Solutions
1. A manufacturer has been selling 1000 flat-screen TVs a week at $450 each. A market
survey indicates that for each $10 rebate offered to the buyer, the number of TVs sold will
increase by 100 per week.
(a) Find the demand function.
(b) How large a rebate should the company offer the buyer in order to maximize its revenue?
(c) If its weekly cost function is C(x) = 68, 000 + 150x, how should the manufacturer set the
size of the rebate in order to maximize its profit?
Solution:
(a) We are given the demand or price function when for 1000 TVs i.e., p(1000) = 450.
If rebate of $10 is given then sale increases by 100 per week. Therefore, for price decrease
1
1
of
× 10 =
per unit, the sale will increase by 100 unit per week.
100
10
Let say x is the number of unit sold per week then x − 1000 is the increase in the number
of units sold.
Thus,
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1
(x − 1000) =⇒ p(x) = 550 − x is the demand function.
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10
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(b) Revenue will be R(x) = xp(x) i.e., R(x) = 550x − x2 =⇒ R0 (x) = 550 − x. R0 (x)
10
5
is called the marginal revenue.
Let’s use First Derivative Test to maximize the revenue:
p(x) = 450 −
1
Critical numbers can be found by R0 (x) = 0 i.e., 550 − x = 0 =⇒ x = 5(550). Therefore
5
critical number is x = 2750 .
Observe that R0 (x) > 0 for 0 ≤ x < 2750 and R0 (x) < 0 for x > 2750. Hence, first
derivative test will lead to the conclusion that maximum occurs at x = 2750.
Now p(2750) = 275 (see equation for demand function in part (a)) thus, the rebate to
maximize the revenue should 450 − 275 = $175 .
(c) We are given C(x) = 68, 000 + 150x and we know profit function P (x) = R(x) − C(x)
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therefore, P (x) = 550x − x2 − 68, 000 + 150x =⇒ P (x) = 400x − x2 − 68000 and
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1
marginal profit is P 0 (x) = 400 − x.
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Let’s use First Derivative Test to maximize the profit:
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1
Critical numbers can be found by P 0 (x) = 0 i.e., 400 − x = 0 =⇒ x = 5(400). Therefore
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critical number is x = 2000 .
Observe that P 0 (x) > 0 for 0 ≤ x < 2000 and R0 (x) < 0 for x > 2000. Hence, first
derivative test will lead to the conclusion that maximum occurs at x = 2000.
Now p(2000) = 350 (see equation for demand function in part (a)) thus, the rebate to
maximize the revenue should 450 − 350 = $100 .
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2. You start a tattoo business. The cost in dollars to tattoo x images in a month is given by
x2
+ 40x + 400, where x ∈ [0, 80].
C(x) =
4
The price of a tattoo is about 70 dollars.
(a) Find the revenue and the profit function.
(b) Find the marginal revenue and the marginal profit.
(c) Compute the marginal revenue and marginal profit if you tattoo 20 images. Explain
what your answer tells you.
(d) Find the number of tattoos you should draw to maximize your profit.
Solution:
(a) Let’s first find the revenue and the profit function.
Revenue function R(x): R(x) = 70x.
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Profit function P (x): P (x) = R(x) − C(x) = 70x − ( x4 + 40x + 400) = 30x −
x2
4
− 400.
(b) Now let’s find the marginal revenue and the marginal profit.
Marginal revenue R0 (x): R0 (x) = (70x)0 = 70.
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Marginal profit P 0 (x): P 0 (x) = (30x − x4 − 400)0 = 30 − x2 .
(c) Next let’s compute the marginal revenue and marginal profit if you tattoo 20 images.
Marginal revenue R0 (20) for x = 20: R0 (20) = 70 and marginal profit P 0 (20) for x = 20:
P 0 (20) = 30 − 20
= 20.
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*Explain what your answer tells you.*
R0 (20) ' R(21) − R(20). This means that the revenue will go up by about 70 dollars if
another tattoo is drawn. P 0 (20) ' P (21) − P (20). This means that the profit will go up by
about 20 dollars if another tattoo is drawn.
(d)
And now let’s find the number of tattoos you should draw to maximize your profit.
We have to examine P (x) in the interval [0, 80] and we can use Closed Interval Test to find
the possible maximum.
We first look at the derivative.
x
P 0 (x) = 30 − = 0 ⇔ x = 60.
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That is x = 60 is the critical point, now observe P (60) = 500, P (0) = −400 and P (80) =
400.
The closed interval test reveals that x = 60 is indeed a maximum. Hence the maximal
profit of your business is 500 dollars a month. It is achieved if you draw 60 tattoos.
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