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lionely

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## Homework Statement

a match box consists of an outer cover, open at both ends, into which slides a rectangular box without a top. The length of the box is one and a half times its breadth, the thickness of the material is negligible, and the volume of the box is 25cm^3 . If the breadth of the box is x cm, find , in terms of x, the area of the material used. Hence show that, if the least area of material is to be used to make the box , the length should be 3.7cm approximately.

## Homework Equations

## The Attempt at a Solution

So I have this so far:

25 = 3/2 x^2 h [Volume and h is the height]

For the surface area now, I figured I would find the surface area of the outer box(box1) and the inner box(box2)

and then add them together.

so I have this :

box1 = ((3x^2)/2 + xh) (2) = 3x^2 + 2xh

box 2 = (3x^2)/2 + xh(2) + (3xh/2)(2) = (3x^2)/2 + 5xh

box1 + box2 = (9x^2)/2 + 7xh [ h = (50)/(3x^2) , from the volume ]

Area(A) = (9x^2)/2 + 350/(3x)

dA/dx = 9x - (350)/(3x^2)

for max/min dA/dx = 0

9x = 350/(3x^2)

x^3 = 350/27

x = 2.45cm (which is incorrect)

Where did I go wrong?

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