Respuesta :
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,
[tex]f(x) = (x-22)(53-x)[/tex]
Where x is the price of the product.
Since for avoiding the loss,
[tex]f(x) \geq 0[/tex],
⇒ [tex](x-22)(53-x)\geq 0[/tex]
If [tex](x-22)\geq 0 \implies x\geq 22[/tex]
If [tex]53 - x \geq 0 \implies 53 \geq x[/tex]
Thus, 53 ≥ x ≥ 22,
Therefore, the minimum price to avoid the loss = $ 22
And, the minimum price to avoid the loss = $ 53
[tex]f'(x) =\frac{d}{dx}[(x-22)(53-x)] = (x-22)\frac{d}{dx} (53-x)+(53-x) \frac{d}{dx}(x-22)[/tex]
⇒ [tex]f'(x) = - (x-22)+(53-x) = -x+22+53-x =-2x+75[/tex]
Now, For maximum or minimum,
[tex]f'(x) = 0[/tex],
⇒ [tex]-2x+75 = 0[/tex]
⇒ [tex]-2x = -75[/tex]
⇒ [tex]x = 37.5[/tex]
[tex]f''(x) = \frac{d}{dx}(-2x+75)=-1[/tex]
For x = 37.5, f''(x) is negative,
Thus, For the price of $ 37.5 dollars the company has the greatest profit.
You can use the fact that loss is denoted by negative profit.
- The minimum price, in dollars. to avoid a loss is 22
- The maximum price, in dollars, to avoid a loss is 53
- The price, in dollars, for which the profit obtained is maximum is 37.5
How to obtain the maximum value of a function?
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.
When does loss happen?
For given function of profit, if the profit goes below 0, then it will denote loss.
Using above conclusions to find the values
Loss is prevented when [tex]f(x) \geq 0[/tex]
Thus, putting value of f(x) in above inequality, we get:
[tex](x-22)(53-x) \geq 0\\x-22 \geq0 \: \rm and \: \: 53-x \geq 0\\x \geq 22 \: and \: 53 \geq x\\or\\22 \leq x \leq 53[/tex]
Thus, to prevent profit, the minimum price in dollars is $22 and the maximum price in dollars is $53
Finding the value of x at which the profit is maximum:
[tex]f(x) = (x-22)(53-x) \\f'(x) = (x-22)(-1) + (53-x)(1) = -2x +75\\f''(x) = -2[/tex]
Getting critical points:
[tex]f'(x) = 0\\-2x + 75 = 0\\\\x = \dfrac{75}{2} = 37.5[/tex]
Since second rate of function of profit is negative for all x, thus the function outputs maximum value at x = 37.5
Thus,
- The minimum price, in dollars. to avoid a loss is 22
- The maximum price, in dollars, to avoid a loss is 53
- The price, in dollars, for which the profit obtained is maximum is 37.5
Learn more about maximum values of functions here:
https://brainly.com/question/13333267
