Maxima word problem - Not sure about my answer.

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The discussion centers on a mathematical optimization problem involving a builder who needs to maximize the volume of a rectangular building with a square base under a budget constraint of $1650. The cost of the walls is $25 per square meter, while the roof costs $55 per square meter. The solution involves deriving the dimensions using the equations for the surface area and volume, ultimately leading to the conclusion that the optimal length for the base is approximately 3.263 meters. The discussion also highlights the importance of calculus in confirming that a cube maximizes volume for a given surface area.

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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) = 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..

================================================
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
================================================
EDIT:

I've established that the Volume would be L2H

V = L2H
A = L2
B = HL
 
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Have you tried finding an expression for the volume in terms of H or L alone?
 


Simon Bridge said:
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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No - I mean you have two simultaneous equations and three unknowns ... why not eliminate one of the unknowns in the equation that starts V= ?

For instance ... from 1650=55L2+100HL ... you can solve it for H, then substitute into V=HL2 ... which gives you V in terms of L. Now you can find the maxima.

You could assume that the shape is a cube - but you still have to prove that a cube is the right shape.
 

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