Question

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs three times as much (per unit area) as glass, use Lagrange multipliers to find the dimensions of the aquarium that minimize the cost of the materials.

Answer 1

Answer:

See details on explanation below...

Explanation:

By having  a value for volume (v) the solution is as followed:

Let's develop 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×d)+(1)(2(w+d)×h)

Minimum will be performed when the width and the depth are the same.

So we are only need the case (substituting w for d)

C=5w^2+4wh

Volume in this case is expressed as

v=w^2h

So if we re-write our Cost equation as

C=5w^2+4w^2h/w

So we obtain

C=5w^2+4vw^−1

By finding the minimum cost, we take the derivative of this and set the result to zero.

dC/dw=10w−4vw^−2

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