Question

A retailer has been selling 1200 tablet computers a week at \(350 each. The marketing department estimates that an additional 80 tablets will sell each week for every \)10 that the price is lowered.

1. Find the demand function.
2. What should the price be set at in order to maximize revenue?
3. If the retailer’s weekly cost function is \(C\left( x \right) = 35,000 + 120x\) what price should it choose in order to maximize its profit?

Short answer

1. The demand function is\(y = p\left( x \right) = - \frac{1}{8}x + 500\), where\(x \ge 1200\).
2. The price is set to maximize revenue is\(\$ 250\).
3. The price should be set to maximize profit is \(\$ 310\).

Step-by-step solution

(a) Step 1: Find the demand function

The demand function\(p\)is linear, just as in example 6. It is given that\(p\left( {1200} \right) = 350\)and concludes that\(p\left( {1280} \right) = 340\)because\(\$ 10\)a reduction in price increase sales about 80 per week.

Obtain the slope for\(p\)as shown below:

\(\begin{aligned}{c}m = \frac{{340 - 350}}{{1280 - 1200}}\\ = - \frac{1}{8}\end{aligned}\)

Obtain the equation of the line as shown below:

\(\begin{aligned}{c}p - 350 = - \frac{1}{8}\left( {x - 1200} \right)\\y = - \frac{x}{8} + \frac{{1200}}{8} + 350\\y = - \frac{1}{8}x + 150 + 350\\y = - \frac{1}{8}x + 500\end{aligned}\)

Thus, the demand function is \(y = p\left( x \right) = - \frac{1}{8}x + 500\), where \(x \ge 1200\).

(b) Step 2: What price should be set to maximize revenue

The revenue is given by\(R\left( x \right) = xp\left( x \right) = - \frac{1}{8}{x^2} + 500x\).

Obtain the derivative of\(R\left( x \right)\)as shown below:

\(\begin{aligned}{c}R'\left( x \right) = \frac{d}{{dx}}\left( { - \frac{1}{8}{x^2} + 500x} \right)\\ = - \frac{{2x}}{8} + 500\\ = - \frac{1}{4}x + 500\end{aligned}\)

When\(R'\left( x \right) = 0\)then,

\(\begin{aligned}{c} - \frac{1}{4}x + 500 = 0\\ - \frac{1}{4}x = - 500\\x = 4\left( {500} \right)\\x = 2000\end{aligned}\)

The price to maximize revenue is shown below:

\(\begin{aligned}{c}p\left( {2000} \right) = - \frac{1}{8}\left( {2000} \right) + 500\\ = - 250 + 500\\ = \$ 250\end{aligned}\)

Thus, the price is set to maximize revenue is\(\$ 250\).

(c) Step 3: What price should a retailer choose to maximize its profit

The cost function is\(C\left( x \right) = 35,000 + 120x\). The profit is shown below:

\(\begin{aligned}{c}P\left( x \right) = R\left( x \right) - C\left( x \right)\\ = - \frac{1}{8}{x^2} + 500x - 35,000 - 120x\\ = - \frac{1}{8}{x^2} + 380x - 35,000\end{aligned}\)

Obtain the derivative of\(P\left( x \right)\)as shown below:

\(\begin{aligned}{c}P'\left( x \right) = \frac{d}{{dx}}\left( { - \frac{1}{8}{x^2} + 380x - 35,000} \right)\\ = - \frac{{2x}}{8} + 380\\ = - \frac{1}{4}x + 380\end{aligned}\)

When\(P'\left( x \right) = 0\)then,

\(\begin{aligned}{c} - \frac{1}{4}x + 380 = 0\\ - \frac{1}{4}x = - 380\\x = 4\left( {380} \right)\\x = 1520\end{aligned}\)

The price to maximize profit is shown below:

\(\begin{aligned}{c}p\left( {1520} \right) = - \frac{1}{8}\left( {1520} \right) + 500\\ = - 190 + 500\\ = \$ 310\end{aligned}\)

Thus, the price should be set to maximize profit is \(\$ 310\).