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Posted by **Mishaka** on Friday, December 16, 2011 at 6:23pm.

- Calculus (Optimization) -
**Damon**, Friday, December 16, 2011 at 6:35pmv = vol = x^2 y

girth = 4 x

length = y

so y + 4x </=108

since maximizing

y + 4x = 108

or

y = 108 - 4x

v = x^2 (108-4x)

v = 108 x^2 - 4 x^3

dv/dx = 216 x - 12 x^2

= 0 for max or min

so

x(216 - 12x) = 0

x = 18 for max

then y = 108 -4(18) = 36

- Calculus (Optimization) -
**Reiny**, Friday, December 16, 2011 at 6:41pmLet the width and the height of the box both be x inches. (That makes the end a square)

Let the length by y inches

distance around lengthwise = 2x + 2y

distance around widthwise = 4x

(think of a ribbon around a box when you wrap for Christmas)

distance = 6x+2y= 108

y = 54 - 3x

I will assume that by "largest box" you mean greatest volume.

V = x^2 y

= x^2(54-3x)

= 54x^2 - 3x^3

dV/dx = 108x - 9x^2 = 0 for a max V

9x^2 = 108x

x = 12

then y = 54 - 36 = 18

the largest box is 18" long, 12" wide, and 12" high.

- Calculus (Optimization) -
**Reiny**, Friday, December 16, 2011 at 6:45pmReading the question again, I think that I took the wrong interpretation and Damon took the right one.

- Calculus (Optimization) -
**Damon**, Friday, December 16, 2011 at 6:58pmIt is just length + x^2

not girth in both directions

http://pe.usps.com/text/qsg300/Q401.htm

- Calculus (Optimization) -
**Damon**, Friday, December 16, 2011 at 6:59pmI mean length + 4x

- Calculus (Optimization) -
**Mishaka**, Friday, December 16, 2011 at 7:03pmBoth of you, thank you very much!!! I arrived at the correct answer width = 18 and length = 36, but I just got that answer by chance and wasn't sure how I could prove (mathematically) that it was indeed correct, your explanations helped tremendously!

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