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Maxima word problem - Not sure about my answer.

  • Thread starter Bradyns
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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
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
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Have you tried finding an expression for the volume in terms of H or L alone?
 
  • #3
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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.
 
Last edited:
  • #4
Simon Bridge
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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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