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Dom, the jolly court jester, needs to create a square-based rectangular box with a volume of 32,000 cubic inches for his king at the lowest possible cost. The cost of material for the sides of the box $0.25 per square inch while the cost of material for the top and bottom of the box cost $1.00 per square inch. What are the dimensions of the box that would minimize of the cost? Show all work, please.

Respuesta :

Answer:  

Dimensions = 20 inches by 20 inches by 80 inches

The floor is 20 inches by 20 inches. The height is 80 inches.

The minimized cost is $2400.

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Work Shown:

h = height of the box, in inches

x = side length of the square base, in inches

x^2 = area of the floor = area of the ceiling

x^2h = volume of the rectangular box

x^2h = 32000

h = 32000/(x^2)

C = total cost in dollars

C = 0.25*(area of the sides) + 1.00*(area of the floor and ceiling)

C = 0.25*(xh+xh+xh+xh) + 1.00*(x^2+x^2)

C = 0.25*4xh + 1.00*2x^2

C = xh + 2x^2

C = x*(32000/(x^2)) + 2x^2

C = (32000/x) + 2x^2

C = (32000/x) + (x*2x^2)/x

C = (32000/x) + (2x^3)/x

C = (32000+2x^3)/x

The goal is to make C the smallest possible, aka we want to minimize it.

Visually we want the lowest point on the cost curve.

We have two options to get this task done:

  1. Use a graphing calculator.
  2. Use calculus (specifically derivatives).

I'll assume your teacher hasn't gone over calculus at this point. I'll go for option 1 mentioned above.

Use a graphing tool like GeoGebra to plot out the cost function curve. See the diagram below. The lowest point on this curve is at (20,2400).  I used the "min" function to determine this lowest point. Keep in mind that x > 0.

This lowest point indicates to us that x = 20 causes C(x) to be the smallest at $2400. This is the minimized cost.

Therefore, the square base should be 20 inches by 20 inches. The height should be:

h = 32000/(x^2) = 32000/(20^2) = 80 inches

The box should be 20 inches by 20 inches by 80 inches.

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Check:

Volume = length*width*height = 20*20*80 = 32000 cubic inches

This helps verify the answer.

Ver imagen jimthompson5910
semsee45

Answer:

Width = 20 inches

Length = 20 inches

Height = 80 inches

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{6.7 cm}\underline{Volume of a square-based rectangular box}\\\\$V=x^2h$\\\\where:\\\phantom{ww} $\bullet$ $x$ is the side length of the base.\\\phantom{ww} $\bullet$ $h$ is the height.\\\end{minipage}}[/tex]

Given the volume of the square-based rectangular box is 32,000 in³, substitute this into the equation and rearrange to isolate h:

[tex]\implies x^2h=32000[/tex]

[tex]\implies h=\dfrac{32000}{x^2}[/tex]

[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Surface Area of a square-based rectangular box}\\\\$S=2x^2+4xh$\\\\where:\\\phantom{ww} $\bullet$ $x$ is the side length of the base.\\\phantom{ww} $\bullet$ $h$ is the height.\\\end{minipage}}[/tex]

Given:

  • Cost of material for the sides of the box = $0.25 per in²
  • Cost of material for the top and bottom of the box = $1.00 per in²

Create an equation for the total cost, C, of the materials based on the equation for the surface area:

[tex]\implies C=(1)2x^2+(0.25)4xh[/tex]

[tex]\implies C=2x^2+xh[/tex]

Substitute the expression for h into the equation for cost to create an equation for C in terms of x:

[tex]\implies C=2x^2+x\left(\dfrac{32000}{x^2}\right)[/tex]

[tex]\implies C=2x^2+\dfrac{32000}{x}[/tex]

[tex]\implies C=2x^2+32000x^{-1}[/tex]

To find the value of x that would minimize the cost, differentiate the equation for cost:

[tex]\implies \dfrac{\text{d}C}{\text{d}x}}=4x-32000x^{-2}[/tex]

[tex]\implies \dfrac{\text{d}C}{\text{d}x}}=4x-\dfrac{32000}{x^{2}}[/tex]

[tex]\implies \dfrac{\text{d}C}{\text{d}x}}=\dfrac{4x^3-32000}{x^{2}}[/tex]

Set the differentiated equation to zero and solve for x:

[tex]\implies \dfrac{4x^3-32000}{x^{2}}=0[/tex]

[tex]\implies 4x^3-32000=0[/tex]

[tex]\implies 4x^3=32000[/tex]

[tex]\implies x^3=8000[/tex]

[tex]\implies x=20[/tex]

Therefore, the side lengths of the base of the box that would minimize the cost are 20 inches.

To find the height of the box that would minimize the cost, substitute the found value of x into the expression for height:

[tex]\implies h=\dfrac{32000}{20^2}[/tex]

[tex]\implies h=\dfrac{32000}{400}[/tex]

[tex]\implies h=80\; \rm in\;(2\:d.p.)[/tex]

Therefore, the dimensions of the box that would minimize cost are:

  • width = 20 inches
  • length = 20 inches
  • height = 80 inches

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