# Thread: Is my set up for Optimization problem correct? very quick!

1. ## Is my set up for Optimization problem correct? very quick!

The problem is below:

Let x = the length of the edge of one cut out corner .

So the objective function is : V = x(10-2x)^2 + x^3

and there wouldn't be a constraint function other than that x >=0 and <= 5

Is this the correct function that i would take derivative and find critical points for?

2. ## Re: Is my set up for Optimization problem correct? very quick!

Originally Posted by lc99
The problem is below:

Let x = the length of the edge of one cut out corner .

So the objective function is : V = x(10-2x)^2 + x^3

and there wouldn't be a constraint function other than that x >=0 and <= 5

Is this the correct function that i would take derivative and find critical points for?
yes, however the constraint is $0 < x < 5$ ... no equality at the endpoints

3. ## Re: Is my set up for Optimization problem correct? very quick!

Sometimes I'm confused cause my professor includes the endpoints so that we could apply the extreme value theory. Cause if we know that end points are 0, the must critical points

4. ## Re: Is my set up for Optimization problem correct? very quick!

on 2nd thought, the usable domain for x would be $0 < x \le 5$

$x = 0$ ... no boxes, therefore zero volume

$x = 5 \implies V = 125$

5. ## Re: Is my set up for Optimization problem correct? very quick!

Wait, how can x be just 5? if it was justs 5, then we dont cut anything??

I'm not sure if my answer is correct if i do it like that...

What i did was:

V = x(10-2x^^2 + x^3
= 5x^3 - 40x^2 +100x

V' = 5(3x-10)(x-20 =0

x = 10/3 , 2

end points are x = 0 , x=5

when x = 0, V = 0
when x =2, V =80
when x=10/3 , V= 2000/27
when x = 5, V = 125

The maximum volume is when x =5 , but doesn't x = 5 mean that there were no squares cut, therefore, you cant really form a paper clip fold with top and bottom not closed? Doesn't the problem ask that there should be both items involved? or not?

6. ## Re: Is my set up for Optimization problem correct? very quick!

Originally Posted by lc99
Wait, how can x be just 5? if it was justs 5, then we dont cut anything??
yes you do ... four 5x5 pieces to make the paper clip "box"