Posted by Mishaka on Friday, December 16, 2011 at 6:23pm.
The U.S. Post Office will accept a box for shipment only if the sum of the length and girth (distance around) is at most 108 inches. Find the dimensions of the largest acceptable box with square ends.
- Calculus (Optimization) - Damon, Friday, December 16, 2011 at 6:35pm
v = vol = x^2 y
girth = 4 x
length = y
so y + 4x </=108
y + 4x = 108
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
x(216 - 12x) = 0
x = 18 for max
then y = 108 -4(18) = 36
- Calculus (Optimization) - Reiny, Friday, December 16, 2011 at 6:41pm
Let 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
= 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 6:45pm
Reading 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 6:58pm
It is just length + x^2
not girth in both directions
- Calculus (Optimization) - Damon, Friday, December 16, 2011 at 6:59pm
I mean length + 4x
- Calculus (Optimization) - Mishaka, Friday, December 16, 2011 at 7:03pm
Both 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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