Chapter 4.7, Problem 23PPSE

Solution

To calculate the smallest amount, that can charge in dollars to make a profit of at least $\$125$ .

The smallest amount the fundraiser can charge for each box is $\$16.34$ .

Given:

The function of the profit of the fundraiser in dollars is:

  $p=-0.5x^2+25x-150$

where

  $x=\text{ price of each box seeds}$ .

Formula Used:

Quadratic formula is given as:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The given function is:

  $p=-0.5x^2+25x-150$

Now, to make a profit of at least $\$125$ , putting $p=125$ and solving for $x$ :

  $\begin{array}{l}125=-0.5x^2+25x-150\\ \Rightarrow 0.5x^2-25x+125+150=0\\ \Rightarrow 0.5x^2-25+275=0\end{array}$

Comparing the above equation with quadratic equation of the form $a{x}^2+bx+c,$ the parameters are:

  $\begin{array}{l}a=0.5\\ b=-25\\ c=275\end{array}$

Using the quadratic formula to find the values of $x$ as:

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \Rightarrow x=\frac{-\left(-25\right)\pm \sqrt{{\left(-25\right)}^2-4\left(0.5\right)\left(275\right)}}{2\left(0.5\right)}\\ \Rightarrow x=\frac{25\pm \sqrt{625-550}}{1}\\ \Rightarrow x=25\pm \sqrt{75}\\ \Rightarrow x=25\pm 5\sqrt{3}\\ \Rightarrow x=5\left(5\pm \sqrt{3}\right)\\ \Rightarrow x=5\left(5+\sqrt{3}\right)\text{ and }x=5\left(5-\sqrt{3}\right)\end{array}$

Next, simplifying the values of $x$ further as:

  $\begin{array}{l}x=5+\left(5+\sqrt{3}\right)\\ \Rightarrow x=5\left(5+1.732\right)\\ \Rightarrow x=5\times 6.732\\ \Rightarrow x=33.66\end{array}$

And

  $\begin{array}{l}x=x=5\left(5-\sqrt{3}\right)\\ \Rightarrow x=5\left(5-1.732\right)\\ \Rightarrow x=5\times 3.268\\ \Rightarrow x=16.34\end{array}$

Among the two values of $x$ , the smallest value is:

  $x=16.34$

Hence, the smallest amount that can be charged is $\$16.34$ for each box.