Question

The manager of a company uses the function shown to model the company’s daily profit based on the price of a product in dollars, x. f(x)=(x-22)(53-x)
What is the minimum price, in dollars, to avoid a loss?
What is the maximum price, in dollars, to avoid a loss?
What is the price, in dollars, that results in the greatest profit?

Answer 1

Answer: 1) The minimum price, in dollars, to avoid a loss = 22,

2) the maximum price in dollars , to avoid a loss = 53,

3) The price that results the maximum profit = 37.5 dollars

Step-by-step explanation:

Since the given function that shows the total profit,

$f(x) = (x-22)(53-x)$

Where x is the price of the product.

Since for avoiding the loss,

$f(x) \geq 0$,

⇒ $(x-22)(53-x)\geq 0$

If $(x-22)\geq 0 \implies x\geq 22$

If $53 - x \geq 0 \implies 53 \geq x$

Thus, 53 ≥ x ≥ 22,

Therefore, the minimum price to avoid the loss = $ 22

And, the minimum price to avoid the loss = $ 53

$f'(x) =\frac{d}{dx}[(x-22)(53-x)] = (x-22)\frac{d}{dx} (53-x)+(53-x) \frac{d}{dx}(x-22)$

⇒ $f'(x) = - (x-22)+(53-x) = -x+22+53-x =-2x+75$

Now, For maximum or minimum,

$f'(x) = 0$,

⇒ $-2x+75 = 0$

⇒ $-2x = -75$

⇒ $x = 37.5$

$f''(x) = \frac{d}{dx}(-2x+75)=-1$

For x = 37.5, f''(x) is negative,

Thus, For the price of $ 37.5 dollars the company has the greatest profit.