Chapter 4, Problem 54RE

To determine

To find : The most profitable number of each kind to make.

Answer to Problem 54RE

The answer:

About $276$ grade A tires and about $552$ grade B tires.

Explanation of Solution

Given information:

The equation is: $y=\frac{40-10x}{5-x}$

Calculation:

Let's call A the profit for 1 grade A tire and BB the profit for 1 grade B tire.

  $A=2B$

So,

  $\begin{array}{c}P=Ax+By\\ =2Bx+By\\ =B(2x+y)\\ y=\frac{40-10x}{5-x}\\ P=B\left(2x+\left(\frac{40-10x}{5-x}\right)\right)\end{array}$

To find $P\text{'}$ where it is $=0$ :

  $\begin{array}{c}{P}^{\prime }=B\left(2+\frac{(5-x)(-10)-(40-10x)(-1)}{{(5-x)}^2}\right.\\ =B\left(2\frac{{(5-x)}^2}{{(5-x)}^2}+\frac{-50+10x+40-10x}{{(5-x)}^2}\right)\\ =B\left(\frac{2\left(25-10x+x^2\right)-10}{{(5-x)}^2}\right)\\ =B\left(\frac{2x^2-20x+40}{{(5-x)}^2}\right)\\ =2B\left(\frac{x^2-10x+20}{{(5-x)}^2}\right)\\ 0=x^2-10x+20\\ \Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-(-10)\pm \sqrt{{(-10)}^2-4(1)(20)}}{2(1)}\\ =\frac{10\pm 2\sqrt{5}}{2}\\ =5\pm \sqrt{5}\\ =5+\sqrt{5}\\ \approx 7.236\\ =5-\sqrt{5}\\ \approx 2.764\end{array}$

Plug it back into the 2nd equation to find $y$ :

  $\begin{array}{c}y=\frac{40-10(2.764}{5-(2.764)}\\ \approx 5.528\end{array}$

Therefore, the required most profitable number of each kind to make:

About $276$ grade A tires and about $552$ grade B tires.