The monthly profit for a small company that makes long-sleeve T-shirts depends on the price per shirt. If the price is too high,sales will drop. If the price is too low, the revenue brought in may not cover the cost to produce the shirts. After months of datacollection, the sales team determines that the monthly profit is approximated byf(p) =-50p +2050p-20,700, where p is theprice per shirt and f(p) is the monthly profit based on that price.(a) Find the price that generates the maximum profit.(b) Find the maximum profit.(c) Find the price(s) that would enable the company to break even. If there is more than one price, use the "and" button.

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Asked Nov 19, 2019
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The monthly profit for a small company that makes long-sleeve T-shirts depends on the price per shirt. If the price is too high,
sales will drop. If the price is too low, the revenue brought in may not cover the cost to produce the shirts. After months of data
collection, the sales team determines that the monthly profit is approximated byf(p) =-50p +2050p-20,700, where p is the
price per shirt and f(p) is the monthly profit based on that price.
(a) Find the price that generates the maximum profit.
(b) Find the maximum profit.
(c) Find the price(s) that would enable the company to break even. If there is more than one price, use the "and" button.
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The monthly profit for a small company that makes long-sleeve T-shirts depends on the price per shirt. If the price is too high, sales will drop. If the price is too low, the revenue brought in may not cover the cost to produce the shirts. After months of data collection, the sales team determines that the monthly profit is approximated byf(p) =-50p +2050p-20,700, where p is the price per shirt and f(p) is the monthly profit based on that price. (a) Find the price that generates the maximum profit. (b) Find the maximum profit. (c) Find the price(s) that would enable the company to break even. If there is more than one price, use the "and" button.

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Expert Answer

Step 1

a)

To maximize the function, the first derivative of the function is set to 0 and the value of p is determined.

Given function is:

Applying derivative with respect to p:

 

The function has extreme value when the derivative is 0.

 

So the function has a maximum value when p = 20.50 i.e. the price that generates the maximum profit is 20.50

f(P) %3-50p* + 2050р - 20700
f'(p)-50(2p)+2050 (1) - 0
S(p) %--100р +2050
S(Р)-0
—100р + 2050 —0
—100р + 2050— 2050- 0—2050
-100р %3D — 2050
-100р
-2050
-100
-100
р3D 20.50
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f(P) %3-50p* + 2050р - 20700 f'(p)-50(2p)+2050 (1) - 0 S(p) %--100р +2050 S(Р)-0 —100р + 2050 —0 —100р + 2050— 2050- 0—2050 -100р %3D — 2050 -100р -2050 -100 -100 р3D 20.50

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Step 2

b)

Plugging p = 20.50 in the function to obtain the maximum value of the function:

So the maximum profit is 312.50

f (p)--50р* + 2050р — 20700
f (20.50) 50(20.50) +2050(20.50) -20700
f (20.50) 50(420.25)+ 2050(20.50)-20700
f (20.50)-21012.5+42025-20700
_
f (20.50) 312.5
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f (p)--50р* + 2050р — 20700 f (20.50) 50(20.50) +2050(20.50) -20700 f (20.50) 50(420.25)+ 2050(20.50)-20700 f (20.50)-21012.5+42025-20700 _ f (20.50) 312.5

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Step 3

c)

For break even, the monthly profit needs to be 0. So plugging f = 0 in the equation and solving for p gives the values of p for which there i...

f(p) 0
-50р* +2050р -20700- 0
-50 р* 2050р, -20700
-0
-50
50
-50
p -41p+414 0
р - 23р- 18р+ 414 %3D0
Р(P-23)-18(р-23) - 0
(р- 23)(р-18) - 0
р-23 %3D0 оr р -18 %3D 0
р323 or p %3D18
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f(p) 0 -50р* +2050р -20700- 0 -50 р* 2050р, -20700 -0 -50 50 -50 p -41p+414 0 р - 23р- 18р+ 414 %3D0 Р(P-23)-18(р-23) - 0 (р- 23)(р-18) - 0 р-23 %3D0 оr р -18 %3D 0 р323 or p %3D18

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