Maximizing container volume by cutting corner squares from sheet metal

[Sombra]

I need a little help. Could someone get me started (or tell me what's wrong with what I have)?

1. Show that if y = x^(-1/2), x>o, then y' + y/2x = 0
First I found the derivative of y, which came to be -1/2 x^(-3/2). Then I added it to y/2x, and substituted the value of y in terms of x, but it didn't work. I'm left with a fraction and nasty exponents.

2. A rectangle is made by cutting out four squares of x cm length from the corners of a 25 cm by 40 cm rectangular sheet of metal and folding the remaining sheet to form the container. What size squares must be cut out in order to maximize the volume of the container?

First, I stated that V= lwh and l= 40-2x, w=25-2x, h= x and, plugging in these values, I found that dV/dx = 1000-260x-12x^2. Then I solved it quadratically and it didn't work.

Could you put me in the right direction? Thanks!

 

[recon]

For Question 1, $$y' + \frac{y}{2x} = 0$$. This means that you have to show that $$y' = -\frac{y}{2x}$$ or $$y' = - \frac{x^{-\frac{1}{2}}}{2x}$$. This is because $$y = x^{-\frac{1}{2}}$$.

You say that you managed to find that $$y' = -\frac{1}{2}x^{-\frac{3}{2}}$$. This is correct.

So now you just have to play around with $$-\frac{1}{2}x^{-\frac{3}{2}}$$ and try to turn it into $$- \frac{x^-{\frac{1}{2}}}{2x}$$.

 

[Curious3141]

I need a little help. Could someone get me started (or tell me what's wrong with what I have)?

1. Show that if y = x^(-1/2), x>o, then y' + y/2x = 0
First I found the derivative of y, which came to be -1/2 x^(-3/2). Then I added it to y/2x, and substituted the value of y in terms of x, but it didn't work. I'm left with a fraction and nasty exponents.

Derivative correct, substitution wrong. Try again, and you should get the right answer.

2. A rectangle is made by cutting out four squares of x cm length from the corners of a 25 cm by 40 cm rectangular sheet of metal and folding the remaining sheet to form the container. What size squares must be cut out in order to maximize the volume of the container?

First, I stated that V= lwh and l= 40-2x, w=25-2x, h= x and, plugging in these values, I found that dV/dx = 1000-260x-12x^2. Then I solved it quadratically and it didn't work.

Could you put me in the right direction? Thanks!

There is a sign wrong in your expression for dV/dx. Fix it and it should be good.

 

[Gokul43201]

I found that dV/dx = 1000-260x-12x^2.

Check your signs there.

Edit : Nevermind. curious just said the same thing.

 

[Sombra]

ok, I got the 1st one. For the second one, I did have the sign right, I just typed it wrong here. I still end up with an irrational quadratic equation.

 

[Gokul43201]

What solution do you get for the quadratic ?

 

[Curious3141]

ok, I got the 1st one. For the second one, I did have the sign right, I just typed it wrong here. I still end up with an irrational quadratic equation.

It's nothing of the sort. I get one clean integer solution, the other rational but nonintegral solution is obviously inadmissible. Check your work again.

 

[recon]

Hmm. One of my solutions was not an integer; it was rational, though.

 

[Gokul43201]

I too get nice numbers. You can solve quite easily by first taking out tha factor of 4 and then writing 65 = 50 + 15.

 

[Curious3141]

Hmm. One of my solutions was not an integer; it was rational, though.

Yeah, I mistyped that post, I've edited it.

 

[Sombra]

hmmm I got 5. That seems to work. Thanks!

 

[recon]

Wait a minute...where did 5 come from?

 

[Gokul43201]

I got 5 too...and 50/3, which is inadmissible.

 

[Curious3141]

5 is right.

 

[recon]

I can barely read my own handwriting...Sorry!