# 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 ->  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

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 Click here to see ALL problems on Surface-area 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(15661)   (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 = reduce fraction h = 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 for h S.A = 3x^2 + 8(x*) S.A = 3x^2 + 8() : S.A = 3x^2 + : Graph this equation and find the min surface area: On a Ti-83, it says, min S.A. at x = 10, which is 900 sq/in, as indicated here. : find h: h = h = 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?