Answer :
Answer:
47.95in by 13.7in by 21.31in
Step-by-step explanation:
Please check my numbers since I am a bit rusty with optimization.
Step One: Find your primary equation.
This equation is from what you are trying to minimize, which is the surface area of a rectangular prism. (SA=2(l*w+l*h+w*h))
Step Two: Find your secondary equation.
This equation is the one you use to eliminate some variables.
In this question, we are given what the volume equals.
V=14,000 cubic inches
V=l*w*h
Step Three: Plug in known data.
We know that V=14,000, 4x=l, 14x=h. Thus, 14,000=(14x)(4x)(y). (I use y instead of width from here on out).
Manipulate this equation to get y by itself. y=[tex]\frac{250}{x^2}[/tex]
Step Four: Plug into primary and simplify.
SA=2(14x)(y)+2(14x)(4x)+2(4x)(y)
When optimizing, we want to get only one variable in the equation.
SA=2(14x)([tex]\frac{250}{x^2}[/tex] )+2(14x)(4x)+2(4x)([tex]\frac{250}{x^2}[/tex])
SA=112x+(7000/x)+(2000/x)
Step Five: Take the derivative and find the critical point.
[tex]\frac{d}{dx}(SA)=224x-(9000/x^2)[/tex]
[tex]0=224x-(9000/x^2)[/tex]
[tex]x=3.425[/tex]
(Check to see if min by plugging in a point before x and after x. Before should be negative and after x should be positive to be a min.)
Step Six: Find dimensions.
Plug in x into the y=[tex]\frac{250}{x^2}[/tex] equation.
y=[tex]\frac{250}{(3.425)^2}[/tex]=21.3113in
Then plug in x into the side lengths 14x and 4x.
14(3.425)=47.95in
4(3.425)=13.7in
Answer:
21.3 in, 13.7 in and 48.0 in
Step-by-step explanation:
To solve this problem, we need to follow these steps:
1. Understand the Given Conditions:
We are given that the volume [tex]\sf V [/tex] of the rectangular prism (box) is 14,000 cubic inches:
[tex]\sf V = l\times w \times h \\\\ = h \times 4x \times 14x [/tex]
[tex] \sf V = 14,000 \textsf{ cubic inches} [/tex]
Equating equation, we get:
[tex]\sf h \times 4x \times 14x = 14,000 \textsf{ cubic inches} [/tex]
Express Volume in Terms of [tex]\sf h [/tex] and [tex]\sf x [/tex]:
Simplify the volume equation:
[tex]\sf 56hx^2 = 14,000 [/tex]
[tex]\sf hx^2 = \dfrac{14,000}{56} = 250 [/tex]
[tex]\sf hx^2 = 250 [/tex]
[tex]\sf h = \dfrac{250}{x^2} [/tex]
Surface Area Formula:
The surface area [tex]\sf A [/tex] of the box is given by:
[tex]\sf A =2(lw + wh + hl) \\\\ = 2(h \times 4x) + 2(h \times 14x) + 2(4x \times 14x) [/tex]
Substitute [tex]\sf h [/tex] from Volume Equation into Surface Area Formula:
Substitute [tex]\sf h = \dfrac{250}{x^2} [/tex] into the surface area formula:
[tex]\sf A = 2 \left( \dfrac{250}{x^2} \times 4x \right) + 2 \left( \dfrac{250}{x^2} \times 14x \right) + 2(4x \times 14x) [/tex]
Simplify Surface Area Formula:
[tex]\sf A = 2 \left( \dfrac{1000}{x} \right) + 2 \left( \dfrac{3500}{x} \right) + 2(56x^2) [/tex]
[tex]\sf A = \dfrac{2000}{x} + \dfrac{7000}{x} + 112x^2 [/tex]
[tex]\sf A = \dfrac{9000}{x} + 112x^2 [/tex]
Find the Value of [tex]\sf x [/tex] to Minimize Surface Area:
To minimize the surface area of the rectangular prism (box), we need to find the value of [tex] x [/tex] that minimizes the surface area function [tex] A(x) = \dfrac{9000}{x} + 112x^2 [/tex].
This can be done by finding the critical points of [tex] A(x) [/tex], which are the points where the derivative [tex] A'(x) [/tex] is equal to zero.
Surface Area Function:
The surface area [tex] A [/tex] of the box in terms of [tex] x [/tex] is given by:
[tex] A(x) = \dfrac{9000}{x} + 112x^2 [/tex]
Find the Derivative [tex] A'(x) [/tex]:
To find the critical points, differentiate [tex] A(x) [/tex] with respect to [tex] x [/tex]:
[tex] A'(x) = \dfrac{d}{dx} \left( \dfrac{9000}{x} + 112x^2 \right) [/tex]
[tex] A'(x) = -\dfrac{9000}{x^2} + 224x [/tex]
Set [tex] A'(x) = 0 [/tex]:
Find the value(s) of [tex] x [/tex] where [tex] A'(x) = 0 [/tex]:
[tex] -\dfrac{9000}{x^2} + 224x = 0 [/tex]
Solve for [tex] x [/tex]:
Rearrange the equation:
[tex] -\dfrac{9000}{x^2} + 224x = 0 [/tex]
Multiply both sides by [tex] x^2 [/tex] to clear the fraction:
[tex] -9000 + 224x^3 = 0 [/tex]
[tex] 224x^3 = 9000 [/tex]
Solve for [tex] x^3 [/tex]:
[tex] x^3 = \dfrac{9000}{224} [/tex]
[tex] x^3 \approx 40.178571428571 [/tex]
Take the cube root of both sides to solve for [tex] x [/tex]:
[tex] x \approx \sqrt[3]{40.1786} [/tex]
[tex] x \approx 3.4250335529722 [/tex]
Calculate [tex]\sf h [/tex] and Final Dimensions:
Substitute [tex]\sf x = 3.4250335529722 [/tex] back into [tex]\sf h = \dfrac{250}{x^2} [/tex] to find [tex]\sf h [/tex]:
[tex]\sf h = \dfrac{250}{(3.4250335529722)^2} [/tex]
[tex]\sf h = \dfrac{250}{11.730854838985} [/tex]
[tex]\sf h \approx 21.311319885160 [/tex]
Therefore, the dimensions of the box that will minimize the surface area while maintaining a volume of 14,000 cubic inches are approximately:
[tex]\sf h \approx 21.311319885160 \\\\ \approx 21.3 \textsf{inches (in nearest tenth)} [/tex]
[tex]\sf 4x \approx 4 \times 3.4250335529722 \\\\ \approx 13.700134211888 \\\\ 13.7 \textsf{inches (in nearest tenth)} [/tex]
[tex]\sf 14x \approx 14 \times 3.4250335529722 \\\\ 47.950469741610 \\\\ \approx 48.0 \textsf{inches (in nearest tenth)} [/tex]
Therefore the dimensions will result in a box with a volume of 14,000 cubic inches and the minimum possible surface area for the given shape is: 21.3 in, 13.7 in and 48.0 in.
