Answer:
The maximum volume of such box is 32m^3
V = x×y×z = 32 m^3
Step-by-step explanation:
Given;
Total surface area S = 48m^2
Volume of a rectangular box V = length×width×height
V = xyz ......1
Total surface area of a rectangular box without a lid is
S = xy + 2xz + 2yz = 48 .....2
To be able to maximize the volume, we need to reduce the number of variables.
Let assume the rectangular box has a square base,that means; length = width
x = y
Substituting y with x in equation 1 and 2;
V = x^2(z) ....3
x^2 + 4xz = 48 .....4
Making z the subject of formula in equation 4
4xz = 48 - x^2
z = (48 - x^2)/4x .......5
To be able to maximize V, we need to reduce the number of variables to 1, by substituting equation 5 into equation 3
V = x^2 × (48 - x^2)/4x
V = (48x - x^3)/4
differentiating V with respect to x;
V' = (48 - 3x^2)/4
At the maximum point V' = 0
V' = (48 - 3x^2)/4 = 0
Solving for x;
3x^2 = 48
x = √(48/3)
x = √(16)
x = 4
Since x = y
y = 4
From equation 5;
z = (48 - x^2)/4x
z = (48 - 4^2)/4(4)
z = 32/16
z = 2
The maximum volume can be derived by substituting x,y,z into equation 1;
V = xyz = 4×4×2 = 32 m^3
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