Chapter 8.3, Problem 26E

To determine

To find :Whether the function has a minimum value or a maximum value.

Answer to Problem 26E

The given function has minimum value $-\frac{17}{4}$ .

Explanation of Solution

Given function: $y=\frac{1}{5}{x}^2-5x+27$

Calculation:

Since the coefficient of $x^2$ is positive, therefore, the given function has minimum value and no maximum value.

To find the minimum value of the function , we will complete the square as follows.

  $\begin{array}{l}y=\frac{1}{5}{x}^2-5x+27\\ y=\frac{1}{5}\left(x^2-25x\right)+27\\ y=\frac{1}{5}\left(x^2-25x+\frac{625}{4}\right)+27-\frac{1}{5}\times \frac{625}{4}\\ y=\frac{1}{5}{\left(x-\frac{25}{2}\right)}^2-\frac{17}{4}\end{array}$

Since,

  $\begin{array}{l}{\left(x-\frac{25}{2}\right)}^2\ge 0\forall x\in R\\ \frac{1}{5}{\left(x-\frac{5}{2}\right)}^2\ge 0\forall x\in R\\ \frac{1}{5}{\left(x-\frac{5}{2}\right)}^2-\frac{17}{4}\ge -\frac{17}{4}\forall x\in R\end{array}$

Therefore, the given function has minimum value $-\frac{17}{4}$ .