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

Algebra ->  Rectangles -> 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       Log On


   

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) About Me  (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 = 88%2FW%5E2
:
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 88%2FW%5E2
W^2 + 3W(88%2FW%5E2) = 82
W^2 + 264%2FW - 82 = 0
:
Plot this equation
+graph%28+300%2C+200%2C+-6%2C+10%2C+-20%2C+20%2C+x%5E2%2B%28264%2Fx%29-82%29+
Using the integer solution: w = 4 ft is the width
then 2(4) = 8 ft is the length
Find the height
h = 88%2F4%5E2
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