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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
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: