Question 541160: from a square sheet of cardboard 40 cm by 40 cm, square corners are cut out so the sides can be folded up to make a box. what dimensions will yield a box of max. volume?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! from a square sheet of cardboard 40 cm by 40 cm, square corners are cut out so the sides can be folded up to make a box. what dimensions will yield a box of max. volume?
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let x = sides (cm) of square corners cut out
40-2x=sides (cm) of bottom of box
(40-2x)^2=area of bottom of box
volume of box=x(40-2x)^2=x(1600-160x+4x^2)
f(x)=4x^3-160x^2+1600x
Normally, I don't do calculus problems, but this one is a max/min problem which requires calculus for an answer. To find the maximum volume, you must take the first derivative of the function, set it equal to zero, then solve for x:
..
f(x)=4x^3-160x^2+1600x
f'(x)=12x^2-320x+1600=0
3x^2-80x+400=0
(3x-20)(x-20)=0
..
3x-20=0
x=20/3
or
x-20=0
x=20
number line for critical points of derivative:
<....+...20/3.....-.....20......+........>
max at x=20/3 and min at x=20
Ans:
To yield a box of max. volume, sides of square cut outs should=20/3 cm
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