At a retail clothing store, to have a weekly demand of q T-shirts, the manager s

Question

At a retail clothing store, to have a weekly demand of q T-shirts, the manager sets the price per shirt to p= -0.25q+26 dollars. It costs the store $5 for each T-shirt they stock and sell.

a) Find a formula for the total weekly profit P(q), as a function of the number of T-shirts q that the store sells each week.

b) Find the number of T-shirts they should sell in order to maximize weekly profit. Use calculus to demonstrate that the quantity you find is a maximum for the weekly profit. Round to the nearest whole number if necessary and include units with your answer.

c) What is the price for a T-shirt, when the weekly profit is at a maximum? Round to two decimal places if necessary and include units with your answer.

Explanation / Answer

1

Revenue=pq=(-0.25q+26)*q=-0.25q^2+26q

Cost=5q

Profit=Revenue-Cost=-0.25q^2+26q-5q=-0.25q^2+21q=q(21-0.25q)

where q is the weekly number of Tshirts sold

2

Profit=(q-0)(21-0.25q)

The above is a quadratic equation

q=0,42 will be the roots of the above quadratic equation

The above equation will be positive in the interval 0-42

And as it is increasing in the range 0 to 42.it will be maximised at 42..So, quantity of Tshirts will be 42

Verifying using calculus

To maximize weekly profit, differentiating profit w.r.t. q, we get

-0.25*2*q+21

Setting it to 0, we get q=21/(0.25*2)=42

differentiaitng again w.r.t. q, we get

-0.25*2

=-0.5

As we see double differentiation is negative, the profit is maximized at q=42 Tshirts

3

p=-0.25q+26

At q=42, price=-0.25*42+26=$15.5

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