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- anonymous

if you have a rectangle that is 10 by 18 and you cut out the corners equally to make it a box how do you maximize the volume and what is the volume?

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- anonymous

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- anonymous

first the expression that gives you the volume of the box
\[V(x)=x(10-x)(18-x)\]
domain will be \((0,5)\)
expand, take the derivative, which will be a quadratic, find the critical points by setting it equal so zero and solve for \(x\)

- anonymous

hope you are good from there, it is basically algebra from here on in

- anonymous

is it a cubic or a quadratic?

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- anonymous

\(V(x)\) is cubic, \(V'(x)\) is degree 2

- anonymous

ooh damn damn damn i am way wrong!!!

- anonymous

oh sorry forgot to take the derivitive. and should it be x(18-2x)(10-2x) bc you take the corner away from each side?

- anonymous

sorry about that
try
\[V(x)=x(10-2x)(18-2x)\] thats better

- anonymous

yes, you are right, i am way off

- anonymous

dont worry about it your the genius

- anonymous

obviously not
but in any case algebra gives you
\[V(x)=4 x^3-56 x^2+180 x\] and then it should be more or less routine

- anonymous

thanks. if you have a box and the lenghth is 12cm and the width is 5cm and the width is increasing at 2 cm/sec and the length is decreasing at 2 cm/sec what is the DA/dt for the area perimeter and diagonal

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