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!