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akotoh · Oct 6, 2009, 04:28 AM

please help me solve problems in maxima and minima.. I still have 5 problems left that I need to solve.. Please help me solve them one by one..

Here's the first problem:
A closed box, whose length is twice its width, is to have a surface of 192 square inches. Find the dimensions of the box when the volume is maximum..

I start it with this..
let width = x
length = 2x

I don't know what to do next.. hope you can help me..

galactus · Oct 6, 2009, 05:15 AM

Since the length is twice the width, we have L=2W.

The surface area is then $S=\underbrace{2(2W^{2})}_{\text{top and bottom}}+ \underbrace{4WH}_{\text{sides}}+\underbrace{2WH}_{ \text{ends}}=4W^{2}+6WH=192$

The volume is $V=2W^{2}H$

Solve S for, say, H and sub into V. It is then in terms of one variable W.

Differentiate, set to 0 and solve for W. L and H will follow.

akotoh · Oct 6, 2009, 06:24 AM

4w^2 + 6wh = 192
h = 192 / 4w^2 + 6wh

then..
V = 2w^2 (192/4w^2+6w)
V= 384w^2 / 4w^2+6w
0 = 384w^2 / 4w^2+6w

is this correct?

Unknown008 · Oct 6, 2009, 07:09 AM

No akotoh!

$4W^2+6WH = 192$

$\cancel{4W^2}+6WH\cancel{-4W^2} = 192-4W^2$

$\frac{\cancel{6W}H}{\cancel{6W}}= \frac{192-4W^2}{6W}$

$H= \frac{192-4W^2}{6W}$

Now continue :)

Zheins · Jan 15, 2011, 10:41 PM

akotoh: your answer is actually wrong.

it should be h=192-4w^2/6wh.

Zheins · Jan 15, 2011, 10:43 PM

I mean 192-4w^2/6w