Chapter 10.2, Problem 6STP

Suppose the height of the freight elevator in your building is half its width w when the doors are all the way open.

a. What is an expression for the maximum side length of a sheet of metal that will fit through the elevator doors?

b. if the height of the elevator is 3 meters, what is the maximum length that will fit through the doors?

To determine

To write an expression for the maximum side length of a sheet of metal that will fit through the elevator doors.

Answer to Problem 6STP

The expression is $d=\frac{w\sqrt{5}}{2}$ .

Explanation of Solution

Given information:

The height of the freight elevator is half its width $w$ .

Calculation:

Let the maximum side length of the sheet of metal be $d$ .

The width of the elevator is $w$ and its height is $\frac{1}{2}w$ .

The expression can be obtained by:

$\begin{array}{l}{d}^2=w^2+{\left(\frac{1}{2}w\right)}^2\\ d^2=w^2+\frac{1}{4}{w}^2\\ d=\sqrt{\frac{5}{4}{w}^2}\\ d=w\sqrt{\frac{5}{4}}\\ d=\frac{w\sqrt{5}}{2}\end{array}$

Hence, the expression is $d=\frac{w\sqrt{5}}{2}$ .

To determine

To find the maximum length that will through the doors.

Answer to Problem 6STP

The maximum length that will pass through the door is $3\sqrt{5}$ meters.

Explanation of Solution

Given information:

The height of the freight elevator is half its width $w$ .

The height of the elevator is $3$ meters.

Calculation :

The width of the elevator is $w$ and its height is $\frac{1}{2}w$ .

Calculate the width:

$\begin{array}{l}\frac{1}{2}w=3\\ w=6\end{array}$

The width is $6$ meters.

The maximum length that will pass through the doors can be calculated by:

$\begin{array}{l}d=\frac{w\sqrt{5}}{2}\\ d=\frac{6\sqrt{5}}{2}\\ d=3\sqrt{5}\end{array}$

Hence, the maximum length that will pass through the door is $3\sqrt{5}$ meters.