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March 28, 2015

March 28, 2015

Posted by **Pat** on Thursday, January 31, 2013 at 9:48pm.

- calculus-max & min -
**Reiny**, Thursday, January 31, 2013 at 10:06pmThe way I read your question:

You have 48 square inches available in total to make the box.

so let the base be x by x inches and the height y inches

total area available = x^2 + 4xy = 48

y = (48 - x^2)/(4x)

volume = x^2 y

= x^2 (48 - x^2)/(4x)

= 12x - x^3/4

d(volume)/dx = 12 - (3/4)x^2

= 0 for a max of volume

(3/4)x^2 = 12

x^2 = 16

x = √16 = 4

if x = 4, then y = (48 - 16)/16 = 2

the box should have a base of 4 inches by 4 inches and a height of 2 inches

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