How do you find the dimensions of the aquarium that minimize the cost of the materials if the base of an aquarium with volume v is made of slate and the sides are made of glass and the slate costs five times as much (per unit area) as glass?

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How do you find the dimensions of the aquarium that minimize the cost of the materials if the base of an aquarium with volume v is made of slate and the sides are made of glass and the slate costs five times as much (per unit area) as glass?

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Answer 1 · 2015-03-06T17:11:23

Without a specific value for the volume, $v$ , only a generalized solution can be developed.

We need to develop an expression for the cost of materials based on one of the dimensions.

Let $w =$ width of the base;
$d =$ depth of the base; and
$h =$ height of the aquarium
$C =$ cost

$C =$ cost of base $+$ cost of walls
$= (5)( w xx d) +(1)( 2(w+d) xx h)$

Note that the minimum will be achieved when the width and the depth are equal. (I think this should be obvious but re-post if it isn't).
So we are only interested in the case (substituting $w$ for $d$ )

$C = 5w^2 + 4wh$

The volume in this case can be expressed as
$v = w^2h$

So if we re-write our Cost equation as
$C = 5w^2 + (4w^2h)/w$

This becomes
$C = 5w^2 + 4vw^(-1)$

To find the minimum cost, we take the derivative of this and set the result to zero.

$(d C)/(dw) = 10w - 4vw^(-2)$

Once a value has been established for the volume ( $v$ ) this equation can be solved for $w$ and via substitution for $d$ and $h$ .

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How do you find the dimensions of the aquarium that minimize the cost of the materials if the base of an aquarium with volume v is made of slate and the sides are made of glass and the slate costs five times as much (per unit area) as glass?

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