[ASK] Minimum Surface Area

[ASK] Minimum Surface Area

Postby Monox D. I-Fly » Mon Nov 23, 2020 12:02 am

The volume of a cuboid box with a square base is 2 litres. The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides. Suppose the side length of its base is x and the height of the cuboid is h. The minimum production cost is reached when the surface area is ....

A. 4 square dm

B. 6 square dm

C. 8 square dm

D. 10 square dm

E. 12 square dm


What I've done:

[tex]x^2t=2→t=\frac{2}{x^2}[/tex]

The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides. I tried to put that into account regarding the surface area f(x), and thus:

[tex]f(x)=2x^2+4(2x)t=2x^2+8x(\frac{2}{x^2})=2x^2+\frac{16}{x}[/tex]

To determine the value of x so that the surface area will be minimum:

f'(x) = 0

[tex]4x-\frac{16}{x^2}=0[/tex]

[tex]4x=\frac{16}{x^2}[/tex]

[tex]4x^3=16[/tex]

[tex]x^3=4[/tex]

[tex]x=\sqrt[3]{4}[/tex]

Minimum surface area =[tex]2(\sqrt[3]{4})^2+\frac{16}{\sqrt[3]{4}}=2\sqrt[3]{16}+\frac{4^2}{4^{\frac13}}=2\sqrt[3]{16}+4^{\frac53}[/tex]


I'm stuck. Pretty sure I misinterpreted the whole "The production cost per unit of its top and its bottom is twice the production cost per unit of its lateral sides." thing. What was I supposed to do?
Monox D. I-Fly
 
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Re: [ASK] Minimum Surface Area

Postby Guest » Mon Nov 23, 2020 9:04 am

What I've done:
[tex]x^2t=2→t=\frac{2}{x^2}[/tex]

I have no idea what this means because I have no idea what "t" is!

The problem gave the size of the box in terms of "x" and "h". Did you just use "t" instead of "h"? That's very confusing!

The top and bottom have area [tex]x^2[/tex]. The four sides each have area [tex]hx[/tex]. It cost twice as much to manufacture the top and bottom as the sides so, taking the cost of the sides to be 1 per unit area, the top and bottom cost 2 per unit area. The total cost is [tex]2(2x^2)+ 4(xh)= 4x^2+ 4xh[/tex]. It looks to me like you have the sides costing twice as much as the top and bottom!

The volume is [tex]x^2h= 2[/tex] so [tex]h= \frac{2}{x^2}[/tex]. We can write the cost as [tex]4x^2+ 4x\left(\frac{2}{x^2}\right)= 4x^2+ \frac{8}{x}[/tex].

To determine the value of x so that the surface area will be minimum:

Why? The problem did not ask about the surface area being a minimum, it asked about cost being minimum!

You want to minimize [tex]4x^2+ \frac{8}{x}[/tex]. The derivative is [tex]8x-\frac{8}{x^2}= 0[/tex].
[tex]x= \frac{1}{x^2}[/tex]
[tex]x^3= 1[/tex]

So x= 1 and h= 2.
Guest
 

Re: [ASK] Minimum Surface Area

Postby Monox D. I-Fly » Mon Nov 23, 2020 11:39 pm

Guest wrote:I have no idea what this means because I have no idea what "t" is!

The problem gave the size of the box in terms of "x" and "h". Did you just use "t" instead of "h"? That's very confusing!

I'm sorry, I unintentionally mixed up the languages here. When I typed the question, I used "h" as height, but when I was copying down my work here, in which the height was symbolized by "t" because in my language "height" is "tinggi", I typed "t" here without realizing.

Anyway, thanks for your help. Glad to finally have found the answer.

Monox D. I-Fly
 
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Joined: Tue May 22, 2018 1:38 am
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