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Solve 821×54 using an area model
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Let x represent the side length of the square end, and let d represent the dimension that is the sum of length and girth. Then the volume V is given by
V = x²(d -4x)
Volume will be maximized when the derivative of V is zero.
dV/dx = 0 = -12x² +2dx
0 = -2x(6x -d)
This has solutions
x = 0, x = d/6
a) The largest possible volume is
(d/6)²(d -4d/6) = 2(d/6)³
= 2(108 in/6)³ = 11,664 in³
b) The dimensions of the package with largest volume are
d/6 = 18 inches square by
d -4d/6 = d/3 = 36 inches long
V = x²(d -4x)
Volume will be maximized when the derivative of V is zero.
dV/dx = 0 = -12x² +2dx
0 = -2x(6x -d)
This has solutions
x = 0, x = d/6
a) The largest possible volume is
(d/6)²(d -4d/6) = 2(d/6)³
= 2(108 in/6)³ = 11,664 in³
b) The dimensions of the package with largest volume are
d/6 = 18 inches square by
d -4d/6 = d/3 = 36 inches long