SOLUTION: Find possible dimensions for a closed box with volume 256 cubic inches, surface area 352 square inches, and length that is twice the width. The width of the box in inches can be_

Algebra ->  Rectangles -> SOLUTION: Find possible dimensions for a closed box with volume 256 cubic inches, surface area 352 square inches, and length that is twice the width. The width of the box in inches can be_      Log On


   

Question 800671: Find possible dimensions for a closed box with volume 256 cubic inches, surface area 352 square inches, and length that is twice the width.
The width of the box in inches can be___.
I started by using the first equation given to help me solve the problem+V=%282w%5E2H%29%2F2W%5E2 then got 128%2Fw%5E2=H so... S=2LW+2WH+2LH...2%282w%29w%2B2w%28128%2Fw%5E2%29%2B2%282w%29%28128%2Fw%5E2%29
Then i got 4w%5E2%2B%28256%2Fw%29%2B%28512%2Fw%29i don't know how to simplify this any more. But, i was told i needed to graph it, then whatever the x-intercepts are becomes the answer.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
V+=L%2AW%2Ah+
given:
V+=256in%5E3
L=2W
so, you have
256in%5E3+=2W%2AW%2Ah+
256in%5E3+=2W%5E2%2Ah+
256in%5E3%2F2W%5E2=h+
128in%5E3%2FW%5E2=h+......height

S=2LW%2B2WH%2B2LH

352in%5E2=2%2A2W%2AW%2B2W%28128in%5E3%2FW%5E2%29%2B2%2A2W%28128in%5E3%2FW%5E2%29

352in%5E2=4W%5E2%2B2%28128in%5E3%2FW%29%2B4%28128in%5E3%2FW%29
352in%5E2=4W%5E2%2B256in%5E3%2FWin%2B512in%5E3%2FWin
352in%5E2%2AW=4W%5E3%2B256in%5E2%2B512in%5E2
352in%5E2%2AW=4W%5E2%2B768in%5E2
4W%5E3-352in%5E2%2AW%2B768in%5E2=0
W%5E3-88in%5E2%2AW%2B192in%5E2=0
%28W-8%29%28W%5E2%2B8W-24%29+=+0


if %28W-8%29++=+0 => W=8in.....one solution
if W%5E2%2B8W-24+=+0 => W+=+%28-8+%2B-+sqrt%28+8%5E2-4%2A1%2A%28-24%29%29%29%2F%282%2A1%29+
W+=+%28-8+%2B-+sqrt%28+64%2B96%29%29%2F2+
W+=+%28-8+%2B-+sqrt%28+160%29%29%2F2+
W+=+%28-8+%2B-+12.65%29%2F2+
other solution:
W+=+%28-8+%2B+12.65%29%2F2+.....we need only positive value since the width cannot be negative value
W+=4.65%2F2+
W+=2.325in+



The width of the box in inches can be: W=8in or W+=2.325in+
if W=8in => L=16in and 128in%5E3%2F64in%5E2=h+=>2in=h+
if W+=2.325in+=> L=4.65inand 128in%5E3%2F5.4in%5E2=h+=>23.68in=h+

+graph%28+600%2C+600%2C+-20%2C+20%2C+-40%2C+10%2C+%28x-8%29%28x%5E2%2B8x-24%29%29+