from a thin piece of cardboard 10 in. by 10in, square corners are cut out so that the sides can be folded up to make a box. what dimensions will yield a box of maximum volume? what is the maximum volume?

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from a thin piece of cardboard 10 in. by 10in, square corners are cut out so that the sides can be folded up to make a box. what dimensions will yield a box of maximum volume? what is the maximum volume?

Calculus1
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each side is \(10-2x\) and the height is \(x\) so the volume is \[V(x)=(10-2x)^2x\]
multiply this mess out , get \[V(x)=4 x^3-40 x^2+100 x\]

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then take the derivative, set it equal to zero, solve for \(x\) to get your answer
you good from there?
ahh I see thank you so much
you get two values of x find the second derivative and see at which point it is negative,that value of x gives maximum volume.
make sure you pick a root that makes sense
don't forget that the domain of your function is only \((0,5)\)
i have an answer if you want me to check yours, let me know
ok so I got 12x^2-80x-100 for the derivative so then I pulled out a gcf 4x(3x-20)=0 and then x=20/3
f(x) = 4x^3 - 40x^2 + 100x f'(x) = 12x^2 - 80x + 100 Check your derivative.
so don't pull out a gcf? im confused
so do I use quadratic equation to solve this problem is stupid
You wrote the derivative with a last term of -100. I'm telling you above that the last term of the derivative is + 100. Now solve 12x^2 - 80x + 100 = 0
Refer to the attachment.
1 Attachment

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