SOLUTION: having trouble setting this up can someone help please A rectangular box with no top (pictured above) is to contain 2250 cubic inches of volume. Find the dimensions of the box t

Algebra ->  Surface-area -> SOLUTION: having trouble setting this up can someone help please A rectangular box with no top (pictured above) is to contain 2250 cubic inches of volume. Find the dimensions of the box t      Log On


   

Question 203465: having trouble setting this up can someone help please
A rectangular box with no top (pictured above) is to contain 2250 cubic inches of volume. Find the dimensions of the box that will minimize the surface area. The length (l) of the base is three times the width (w).




Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangular box with no top (pictured above) is to contain 2250 cubic inches of volume.
Find the dimensions of the box that will minimize the surface area.
The length (l) of the base is three times the width (w).
:
But what about the height of the box? Is it the same as the width,
or do we have to find it? (h), assuming this is the case:
:
l * w * h = 2250 cu/inches
Let w = x, then we can write it:
3x * x * h = 2250
3x^2 * h = 2250
h = 2250%2F%283x%5E2%29
reduce fraction
h = 750%2Fx%5E2 is the height of the box
:
Surface area for box without a top
S.A. = (l*w) + 2(l*h) + 2(w*h)
:
Substitute 3x for l and x for w
S.A. = 3x*x + 2(3x*h) + 2(x*h)
S.A = 3x^2 + 6hx + 2hx
S.A = 3x^2 + 8hx
Substitute 750%2Fx%5E2 for h
S.A = 3x^2 + 8(x*750%2Fx%5E2)
S.A = 3x^2 + 8(750%2Fx)
:
S.A = 3x^2 + 6000%2Fx
:
Graph this equation and find the min surface area:
+graph%28+300%2C+200%2C+-6%2C+20%2C+-500%2C+2000%2C+3x%5E2+%2B+%286000%2Fx%29%29+
On a Ti-83, it says, min S.A. at x = 10, which is 900 sq/in, as indicated here.
:
find h:
h = 750%2Fx%5E2
h = 750%2F100
h = 7.5
Then l= 30, w=10, h=7.5; the dimensions of the box
:
Check volume: 30 * 10 * 7.5 = 2250
:
Did this make sense?