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A company wishes to construct a rectangular closed storage tank with a square (horizontal) base. The tank must have a fixed volume of 100 cubic metres. Find the dimensions for the minimum cost of material;

a. If metal for sides and top costs $1.25/square metre and metal for base costs

$4.75/square metre.

b. Repeat (a) if the 12 edges must be welded at a cost of $7.50/metre of weld.

So, I tried by doing this:

l=length w=width h=height v=volume

v=lxwxh Since l=2w, we have v=2wxwh ie 100=2hw^2

That also implies that h=50/w^2

I am having doubts on how to continue because it says the storage is rectangular but the base is squared. I'm not sure how that's possible.

William
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  • kindly include your attempt. – Siong Thye Goh Nov 23 '17 at 21:19
  • Let the side of the base be $s$ and the height be $h$. Can you write a formula for the area of each side? Can you write a formula for the cost? What have you tried? – Ross Millikan Nov 23 '17 at 21:25
  • As you don't comment in any manner the homework you have, we could think that you want a "ready-for-use" solution. But surely not because you have worked hard on the question: thus say us where you are blocked.... – Jean Marie Nov 23 '17 at 21:29
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    I tried adding the area of each side including the base and top, and multiplied it by the variou sprices, but I'm not too sure about it, since it concerns a "closed" box. I'll edit the question and put up my solution in just a minute... – William Nov 23 '17 at 21:30
  • The base is given to be square, so the length and width are the same. – Ross Millikan Nov 23 '17 at 22:23

2 Answers2

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  • $b$ = base side length in meters (only one since it is square).
  • $h$ = height length in meters.

$a)$

$$ \min_{b,h} 1.25(2b^2) + 4.75(4bh) \quad\text{subject to}\quad b^2h=100. $$

$b)$

$$ \min_{b,h} 1.25(2b^2) + 4.75(4bh) + 7.5(8b+4h) \quad\text{subject to}\quad b^2h=100. $$

Should be straight-forward from there.

mzp
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  • It is easier to use $b^2h=100$ to eliminate one of the variables, say substitute $h=\frac {100}{b^2}$ and work in only one variable. – Ross Millikan Nov 23 '17 at 22:25
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Let $l$ and $w$ denote the length and width of the base and $h$ the height of the tank, respectively. Since the base is square, we have $l = w$. Also, the volume is give by $$lwh = l^2 h =100. \tag 1$$ Now, for the first part, the cost function is $$C = 4.75 l^2 + 1.25(4lh + l^2) = 6 l^2 + 5 lh. \tag{2}$$ Now use $(1)$ to compute $C$ as a function of one variable (say $l$) and minimize $C$ by computing $dC/dl$.

The procedure for the second part is very similar to the first part. I leave the calculations to you.

Math Lover
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