Mathematics, 30.03.2021 02:50 mirandaperez3412

A rectangular box is to have a square base and a volume of 40 ft3. If the material for the base costs $0.34 per square foot, the material for the sides costs $0.05 per square foot, and the material for the top costs $0.16 per square foot, determine the dimensions of the box that can be constructed at minimum cost.

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Answer from: keely8468

The dimensions of the box so that total costs are minimum are a side length of 2 feet and a height of 5 feet.

Step-by-step explanation:

Geometrically speaking, the volume of the rectangular box (), in cubic feet, is represented by this formula:

$V = l^{2}\cdot h$ (1)

Where:

- Side length of the box, in feet.

- Height of the box, in feet.

In addition, the total cost of the box (), in monetary units, is defined by this formula:

$C = (c_{b}+c_{t})\cdot l^{2} + 4\cdot c_{s}\cdot l\cdot h$ (2)

Where:

$c_{b}$ - Unit cost of the base of the box, in monetary units per square foot.

$c_{t}$ - Unit cost of the top of the box, in monetary units per square foot.

$c_{s}$ - Unit cost of the side of the box, in monetary units per square foot.

By (1), we clear into the expression:

$h = \frac{V}{l^{2}}$

And we expand (2) and simplify the resulting expression:

$C = (c_{b}+c_{t})\cdot l^{2}+4\cdot c_{s}\cdot \left(\frac{V}{l} \right)$ (3)

If we know that $c_{b} = 0.34\,\frac{m.u.}{ft^{2}}$ , $c_{s} = 0.05\,\frac{m.u.}{ft^{2}}$ , $c_{t} = 0.16\,\frac{m.u.}{ft^{2}}$ and $V = 40\,ft^{3}$ , then we have the resulting expression and find the critical values associated with the side length of the base:

$C = 0.5\cdot l^{2} + \frac{8}{l}$

The first and second derivatives of this expression are, respectively:

$C' = l -\frac{8}{l^{2}}$ (4)

$C'' = 1 + \frac{16}{l^{3}}$ (5)

After equalizing (4) to zero, we solve for : (First Derivative Test)

$l-\frac{8}{l^{2}} = 0$

$l^{3}-8 = 0$

$l = 2\,ft$

Then, we evaluate (5) at the value calculated above: (Second Derivative Test)

C'' = 3

Which means that critical value is associated with minimum possible total costs. By (1) we have the height of the box:

$h = 5\,ft$

The dimensions of the box so that total costs are minimum are a side length of 2 feet and a height of 5 feet.

Answer from: Quest

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step-by-step explanation:

Answer from: Quest

$w\geq 14$

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

$2l+2w\geq 40$

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

$2\times 6+2w\geq 40$

now, we can solve for w

$12+2w\geq 40$

subtract both sides by 12

$12+2w-12\geq 40-12$

$2w\geq 28$

divide both sides by 2

and we get

$w\geq 14$

so,

the smallest possible measurement of the width is 14 feet

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