What is the optimal volume of a box with given dimensions using calculus?

Elihu5991
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Homework Statement


SEE QUESTION IMAGE


Homework Equations


SEE ABOVE


The Attempt at a Solution


SEE WORKINGS IMAGE
 

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Can you type in your work?

ehild
 
Sorry that my handwritiing is too messy and scan isn't done right. Just trying to get this done ASAP.

(25-2x)(40-2x)

y=(25-2x)(-2)+(40-2x)(-2)
=-50+4x-40+2x
=-90+6x
x=15

Not sure what to next do.
 
Elihu5991 said:
Sorry that my handwritiing is too messy and scan isn't done right. Just trying to get this done ASAP.

(25-2x)(40-2x)

y=(25-2x)(-2)+(40-2x)(-2)
=-50+4x-40+2x
=-90+6x
x=15

Not sure what to next do.

Is x the length of the corner pieces cut out? If so, you need to define it in words; a formulation without an explanation is worth 0.

Also, is (25-2x)*(40-2x) the thing you want to maximize? Why? What does the box look like if you use your proposed solution of x = 15?
 
Elihu5991 said:
Sorry that my handwritiing is too messy and scan isn't done right. Just trying to get this done ASAP.

(25-2x)(40-2x)

y=(25-2x)(-2)+(40-2x)(-2)
You mean y', not y.

Also, I'm guessing that you want to maximize the volume. If so, then
y = (25-2x)(40-2x)
as the volume equation would be wrong. You're missing the height. What would it be?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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