Chapter 2.1, Problem 81E

To determine

To find:The value of b such that the function has the given maximum value.

Answer to Problem 81E

The required value of b is $b=\pm 20$

Explanation of Solution

Given data:

The function is $f\left(x\right)=-x^2+bx-75$ and the given maximum value $f_{\max .}\left(x\right)=25$

Concept used:

The standard form of quadratic function $f\left(x\right)=a{\left(x-h\right)}^2+k$ represents a parabola with vertex $\left(h,k\right)$ . If $a<0$ then the graph of quadratic function opens downwards, and maximum value of function is at the vertex, i.e. vertex is $\left(h,k\right)=\left(h,f_{\max .}\right)$

The givenfunction is $f\left(x\right)=-x^2+bx-75$ and $f_{\max .}\left(x\right)=25$ .

The function $f\left(x\right)$ may be written as

  $\Rightarrow f\left(x\right)=-\left(x^2-2\cdot \frac{b}{2}x+\frac{b^2}{4}-\frac{b^2}{4}\right)-75$

  $\Rightarrow f\left(x\right)=-\left(x^2-2\cdot \frac{b}{2}x+\frac{b^2}{4}\right)+\frac{b^2}{4}-75$

  $\Rightarrow f\left(x\right)=-1\cdot {\left(x-\frac{b}{2}\right)}^2+\frac{b^2}{4}-75$...............(1)

This is standard form of the quadratic function $f\left(x\right)$ . Now, comparing result (1) with $f\left(x\right)=a{\left(x-h\right)}^2+k$ . We have $a=-1$ , $h=\frac{b}{2}$ , and $k=\frac{b^2}{4}-75$ .

Since, $a=-1<0$ , so the function $f\left(x\right)$ concave downward and so its vertex is

  $\left(h,k\right)=\left(\frac{b}{2},\frac{b^2}{4}-75\right)=\left(h,f_{\max .}\right)$

  $\Rightarrow f_{\max .}=\frac{b^2}{4}-75$ .

But $f_{\max .}\left(x\right)=25$ , therefore $\frac{b^2}{4}-75=25$ .

  $\Rightarrow b^2=4\left(25+75\right)=400$

  $\Rightarrow b=\pm \sqrt{400}=\pm 20$

Thus, the function $f\left(x\right)=-x^2+bx-75$ has maximum value $f_{\max .}\left(x\right)=25$ for $b=\pm 20$ .

The answer can be verified from figure given below.

  

Thus, for value $b=\pm 20$ the function $f\left(x\right)=-x^2+bx-75$ has maximum value 25.

Conclusion:

For $b=\pm 20$ the function $f\left(x\right)=-x^2+bx-75$ has maximum value 25.