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
:
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/W^2}}}
:
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/W^2}}}
W^2 + 3W({{{88/W^2}}}) = 82
W^2 + {{{264/W}}} - 82 = 0
:
Plot this equation
{{{ graph( 300, 200, -6, 10, -20, 20, x^2+(264/x)-82) }}}
Using the integer solution: w = 4 ft is the width
then 2(4) = 8 ft is the length
Find the height
h = {{{88/4^2}}}
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