Chapter 9, Problem 10MPS

To determine

To Find: The original dimension of the receiver.

Answer to Problem 10MPS

  $\frac{22}{3}\text{ in by }\frac{44}{3}\text{ in}$

Explanation of Solution

Given: The rectangular base of a radio receiver is twice as long as it is wide. A design engineer finds that in order to fit the receiver into the space available on an airplane, she must decrease the length by 7 in. and increase the width by 2 in, If this changes decrease the area of the base by 32 ${\text{in}}^2$ .

Let the width of base be $x\text{ in}$

Length be $2x\text{ in}$

A design engineer wants to fit receiver into the space available on an airplane.

So, decrease the length by 7 in and increase the width by 2 in.

New length $=\left(2x-7\right)\text{ in}$

New width $=\left(x+2\right)\text{ in}$

Original area $=2x^2{\text{ in}}^2$

New area $=\left(2x-7\right)\left(x+2\right){\text{ in}}^2$

Change in area $=36{\text{ in}}^2$

  $\begin{array}{l}\therefore 2x^2-\left(2x-7\right)\left(x+2\right)=36\\ 2x^2-2x^2+7x-4x+14=36\\ 3x=36-14\\ x=\frac{22}{3}\end{array}$

Original dimension of radio receiver $\frac{22}{3}\text{ in by }\frac{44}{3}\text{ in}$