Question

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise A rectangular box which is open at the top can be made from an 24-by-30-inch piece of metal by cutting a square from each corner and bending up the sides. Find the dimensions of the box with greatest volume, where h = height, l = length, and w = width. (Note: let the width be determined by the 24-inch side and the length by the 30-inch side.)

Answer 1

Answer:

The maximum volume is 1417.87 $inch^3$

Explanation:

Optimization Using Derivatives

We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches

When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

$V=(24-2x)(30-2x)x$

Operating

$V=4x^3-108x^2+720x$

To find the maximum value of V, we compute the first derivative and equate it to zero

$V'=12x^2-216x+720=0$

Simplifying by 12

$x^2-18x+60=0$

Completing squares

$x^2-18x+81-81+60=0$

$(x-9)^2=21$

We have two values for x

$x=9+\sqrt{21}=13.58\ inch$

$x=9-\sqrt{21}=4.42\ inch$

The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16

We'll keep only the solution

$x=4.42\ inch$

The width is

$w=(24-2(4.42))=15.16\ inch$

The length is

$l=(30-2(4.42))=21.16\ inch$

And the height

$x=4.42\ inch$

The maximum volume is

$V=(15.16)(21.16)(4.42)=1417.87\ inch^3$