Word Problem: Minimizing Surface Area

tarynt1

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A rectangular box with a closed top is to be made with a volume of 6912 cubic inches. The length of the box must be three times its width. What are the dimensions that would use the least amount of material?

In other words, I need the values of the length, width, and height of the box that will minimize its surface area. ..Right?

I have a diagram of a 3D rectangular box, with the width being x, the length being 3x, and the height being y.

The primary equation is the surface area of the box, (SA) = 8xy + 6y^2.

The secondary equation is for volume, V = 3(x^2)y, or 6912 = 3(x^2)y.

I started byisolating y in the volume equation: y = (6912)/(3x^2). It simplifies to y = (2304)/(x^2).

I substituted the new y value in the surface area formula to get SA = (8x)(2304/x^2) + 6x^2. After simplifying it is SA = 18432/x + 6x^2.

I took the derivative of the surface area formula to get dSA/dx = 18432/x^2 + 12x = 0.

The next step is to find the critical numbers, which I am having trouble finding. It's very possible that I've done this entire problem wrong since I only recently learned this and quite possibly have no idea what I'm doing.

I'd appreciate any help. Thanks!
 
A rectangular box with a closed top is to be made with a volume of 6912 cubic inches. The length of the box must be three times its width. What are the dimensions that would use the least amount of material?

In other words, I need the values of the length, width, and height of the box that will minimize its surface area. ..Right?

I have a diagram of a 3D rectangular box, with the width being x, the length being 3x, and the height being y.

3x<sup>2</sup>y = 6912
x<sup>2</sup>y = 2304
y = 2304/x<sup>2</sup>

surface area = 6x<sup>2</sup> + 8xy

S = 6x<sup>2</sup> + 8x(2304/x<sup>2</sup>)

S = 6x<sup>2</sup> + 18432/x

dS/dx = 12x - 18432/x<sup>2</sup>

12x - 18432/x<sup>2</sup> = 0

12x = 18432/x<sup>2</sup>

x<sup>3</sup> = 1536

x = 8(3)<sup>1/3</sup>
 
You have:

. . . . .width: x
. . . . .length: 3x
. . . . .height: y

You are given the volume, so:

. . . . .V = Lwh

. . . . .6912 = (3x)(x)(y)

. . . . .6912 = 3x<sup>2</sup>y

. . . . .2304/x<sup>2</sup> = y

This gives you the height y in terms of the variable x. Then the surface area SA is given by:

. . . . .SA = 2(x)(3x) + 2(x)(y) + 2(3x)(y)

. . . . .SA = 6x<sup>2</sup> + 4608/x + 13824/x

. . . . .SA = 6x<sup>2</sup> + 18432/x

. . . . .SA = 6x<sup>2</sup> + 18432x<sup>-1</sup>

To find the critical numbers, differentiate and set equal to zero:

. . . . .d(SA)/dx = 12x - 18432x<sup>-2</sup>

The only difference appears to be that you lost a "minus" sign somewhere. Make that correction, and you should be able to complete the exercise.

Eliz.

P.S. Thank you for showing your work! :D
 
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