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**Maxima word problem -- Not sure about my answer.**

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) = L

^{2}

Walls (B) = 4 × (L

^{2})

$1650 = $55A + $25(4B)

$1650 = $55A + $100B

$1650 = $55L

^{2}+ $100L

^{2}

$1650 = $155L

^{2}

$1650 / $155 = L

^{2}

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..

================================================

Roof (A) = L

^{2}

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

1650 = A + 4B

1650 = L

^{2}+ 4HL

1650 = 55L

^{2}+ 100HL

55L

^{2}+ 100HL -1650 = 0

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

Also, using

**1650 = 55L**,

^{2}+ 100HLI managed to arbitrarily plug and chug a result of L=3m H=3.85m

================================================

EDIT:

I've established that the Volume would be

**L**

^{2}H**V = L**

^{2}H**A = L**

^{2}**B = HL**

Last edited: