SOLUTION: solve the problem using a system of two equations in two unknowns. a chief engineer built a tank of transparent aluminum to hold a pair of cetaceans. if the 25-ft-high tank had a v

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Question 930240: solve the problem using a system of two equations in two unknowns. a chief engineer built a tank of transparent aluminum to hold a pair of cetaceans. if the 25-ft-high tank had a volume of 37,500 ft^3 and it took 7000ft^2 of transparent aluminum to cover all six sides, then what were the length and width of the tank
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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solve the problem using a system of two equations in two unknowns.
a chief engineer built a tank of transparent aluminum to hold a pair of cetaceans.
if the 25-ft-high tank had a volume of 37,500 ft^3 and it took 7000ft^2 of transparent aluminum to cover all six sides, then what were the length and width of the tank
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Find LW using the volume
LW * 25 = 37500
LW = 37500/25
LW = 1500 sq/ft, the area of the base and the top
Also
L = 1500%2FW
:
Surface area:
2(LW) + 2(LH) + 2(WH) = 7000
Simplify divide by 2
LW + LH + WH = 3500
Replace H with 25, Replace LW with 1500
LW + 25L + 25W = 3500
1500 + 25L + 25W = 3500
25L + 25W = 3500 - 1500
25L + 25W = 2000
simplify divide by 25
L + W = 80
replace L with 1500/W
1500%2FW + W = 80
multiply by W
1500 + W^2 = 80W
A quadratic equation
W^2 - 80W + 1500 = 0
Factors to
(W-30)(W-50) = 0
Two solutions, the length and width of the tank
W = 30, then L = 50
And
W = 50, then L = 30
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