This was in an assignment handed out after our introduction to extrema's.
A builder wants to build a small building consisting of 4 walls and a roof (all rectangular). The cost of the bricks for the walls will be $25/square metre, while the roof materials will cost$55/square metre. The building must have a square base, but otherwise may have any dimensions the builder wishes. If the builder’s budget is $1650 then what dimensions would make the build have the largest possible volume? The attempt at a solution Premise: A cube will maximize the volume for surface area in this solution. Roof (A) = L2 Walls (B) = 4 × (L2)$1650 = $55A +$25(4B)
$1650 =$55A + $100B$1650 = $55L2 +$100L2
$1650 =$155L2
$1650 /$155 = L2

L = √($1650 /$155)

L≈3.262692338m
L≈3.263m (3.d.p)

For a calculus problem, I am scared that I have not had to use calculus. :/
There was another approaches before I used this method..
Such as..

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Roof (A) = L2
Walls (B) = 4 × (HL) <------------Where h is an variable height.

1650 = A + 4B
1650 = L2 + 4HL
1650 = 55L2 + 100HL

55L2 + 100HL -1650 = 0

But, I couldn't think beyond this.. :/

Also, using 1650 = 55L2 + 100HL,
I managed to arbitrarily plug and chug a result of L=3m H=3.85m
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EDIT:

I've established that the Volume would be L2H

V = L2H
A = L2
B = HL

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Homework Helper

Have you tried finding an expression for the volume in terms of H or L alone?

Have you tried finding an expression for the volume in terms of H or L alone?

In which way?

I could make V= 55L2H
{Area of the base × height.}

I haven't tried using my premise that the most efficient {volume:surface area} is a cube, with the implication of V= L3

I'll give that a shot.

Otherwise, I am truly stumped.

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