Answer:
(2, 2, 1)
Step-by-step explanation:
A squared based will yield the least amount of cardboard for the box. Thus, the volume of the box is:
[tex]V=x^2*h\\h=\frac{4}{x^2}[/tex]
Where 'x' is the length of the sides of the base and 'h' is the height of the box.
The total area of cardboard used is:
[tex]A=x^2+4xh\\A=x^2+4x*\frac{4}{x^2}\\A=x^2+\frac{16}{x}[/tex]
The value of 'x' for which the derivate of the area function is zero, is the base dimension that will yield the least amount of cardboard:
[tex]A=x^2+\frac{16}{x}\\A'=2x-\frac{16}{x^2} =0\\x^3=8\\x=2\ ft[/tex]
The height of the box is then:
[tex]h=\frac{4}{2^2}\\h=1\ ft[/tex]
The dimensions of the box are 2 ft x 2 ft x 1 ft (2, 2, 1).
The dimensions of the box (in feet) that require the least amount of cardboard are; (2, 2, 1).
Thus;
Length = width = x
Let the height be h
Formula for volume of box is;
V = x²h
We are told that volume is 4 ft³. Thus;
4 = x²h
h = 4/x²
Formula for surface area of the open cube is;
S = x² + 4hx
Put 4/x² for h to get;
S = x² + 4x(4/x²)
S = x² + 16/x
Thus;
S' = 2x - 16/x²
At S' = 0;
2x - 16/x² = 0
2x = 16/x²
x³ = 8
x = ∛8
x = 2 ft
Thus;
h = 4/x² = 4/2²
h = 4/4
h = 1 ft
In conclusion, the dimensions of the box (in feet) that require the least amount of cardboard are; (2, 2, 1).
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