Answer: height (h) = 1.59ft
length (x) = 1.59ft
width = 1.59ft
Total cost is $25.28 at these dimensions
Step-by-step explanation:
The volume of a box is given by multiplying its length by its width and height.
If x = length
h = height
it's safe to assume the box is a cube (having a square bottom): length = height = width
Volume = length x width x height = x³
Volume = x³
4 ft² = x³
x = 4^(1/3)
x = 1.587 =1.59ft approximately
Area of square bottom = 1.59 x 1.59 = 2.528ft²
Cost of square bottom = 2.528 x $7 (per ft²) = $17.698
Area of side = 2.528ft²
Cost of side = 2.528 x $3 (per ft²) = $7.584
Total cost = $17.698 + $7.584 = $25.28 to two decimal places.
Answer:
Dimension of the box :
[tex]x =1.19[/tex] ft and [tex]h = 2.82[/tex] ft
Step-by-step explanation:
This is a typical optimization problem.There is a cost function ,A(x,h) and a constraint condition
[tex]4 = x^2\times h[/tex]
[tex]\rm A(x,h) = 2\times x^2\times 7 + 4\times x\times h \times 3[/tex]
The 2 is there because there are two surfaces costing $7/[tex]\rm ft^2[/tex] and the 4 is there because there are four sides costing $3/[tex]\rm ft^2[/tex] .
The way ahead is to solve the constraint condition for h
[tex]h = \dfrac {4}{x^2}[/tex]
substitute for h in A, to get
[tex]\rm A = 14 \times x^2 + \dfrac {48}{x}[/tex]
Now A depends only on x, so the derivative of A with respect to x can be calculated and set equal to zero.
Resulting equation can be solved for x .
[tex]28x -\dfrac{48}{x^2}=0[/tex]
[tex]x^3 = 1.71428571[/tex]
[tex]x=1.19681696[/tex]
[tex]x = 1.19[/tex] (upto two decimal places)
Plugging this back into the constraint condition gives
[tex]h=\dfrac{4}{1.19^2} = 2.82 \;(\rm upto\; two \;decimal \;places)[/tex]
Therefore, dimension of the box is
[tex]x=1.19[/tex] ft and [tex]h=2.82[/tex] ft
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