W=zr-2zt^3 solve for z
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Answer:
Maximum volume of the box will be 7.41 cubic feet.
Step-by-step explanation:
Open box has been made from a metal sheet measuring 3 ft and 8 ft.
Let four square pieces were removed from the four corners with one side measurement x ft.
Volume of the open box = Length × width × height
Length of the box = (3 - 2x) ft
Width of the box = (8 - 2x) ft
Height of the box = x ft
Volume of the box = (3 - 2x)(8 - 2x)x
V = (24 - 6x - 16x + 4x²)x
V = 24x - 22x² + 4x³
Now we take the derivative of V with respect to x and equate the derivative to zero,
V' = 24 - 44x + 12x²
V' = 0
12x² - 44x + 24 = 0
3x² - 11x + 6 = 0
3x² - 9x - 2x + 6 = 0
3x(x - 3) - 2(x - 3) = 0
(3x - 2)(x - 3) = 0
(3x - 2) = 0
and (x - 3) = 0
Therefore, x = 3,
For x = 0.67
Length of the box = (3 - 2x) = 3 - 1.34
= 1.66 ft
Width of the box = (8 - 2x) = 8 - 1.34
= 6.66 ft
Volume of the box = 0.67 × 1.66 × 6.66
V = 7.41 cubic feet.
Similarly, for x = 3,
Length of the box = (3 - 2\times 3) = -3
which is negative but the length of the box can not be negative.
Therefore, maximum volume of the box will be 7.41 cubic feet.