SOLUTION: find the dimensions of an open box whose base is a rectangle whose length is twice its width. if its volume is 176 cubic feet and the totals surface area is 164 square feet

Question 952435: find the dimensions of an open box whose base is a rectangle whose length is twice its width. if its volume is 176 cubic feet and the totals surface area is 164 square feet

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find the dimensions of an open box whose base is a rectangle whose length is twice its width.
if its volume is 176 cubic feet and the totals surface area is 164 square feet
:
Vol: L * W * h = 176
and
S.A.: (L*W) + 2(L*h) + 2(W*h) = 164
:
"length is twice its width."
L = 2W
Replace L with 2W
Vol: 2W * W * h = 176
2W^2*h = 176
Simplif, divide by 2
W^2h = 88
h =
:
S.A.: 2W^2 + 2(2W*h) + 2(W*h) = 164
Simplify,divide by 2
W^2 + 2Wh + Wh = 82
W^2 + 3Wh = 82
replace h with
W^2 + 3W() = 82
W^2 + - 82 = 0
:
Plot this equation

Using the integer solution: w = 4 ft is the width
then 2(4) = 8 ft is the length
Find the height
h =
h = 5.5 ft is the height
:
:
Check this, find the surface area with these dimensions
(8*4) + 2(8*5.5) + 2(4*5.5) =
32 + 88 + 44 = 164

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