# Homework Help: Solve integral quadratically

1. Jan 11, 2005

### 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 derivitive 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!

2. Jan 11, 2005

### 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 $$- \frac{x^-{\frac{1}{2}}}{2x}$$.

Last edited: Jan 11, 2005
3. Jan 11, 2005

### Curious3141

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

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

4. Jan 11, 2005

### Gokul43201

Staff Emeritus

Edit : Nevermind. curious just said the same thing.

5. Jan 11, 2005

### 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.

6. Jan 11, 2005

### Gokul43201

Staff Emeritus
What solution do you get for the quadratic ?

7. Jan 11, 2005

### Curious3141

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.

Last edited: Jan 11, 2005
8. Jan 11, 2005

### recon

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

9. Jan 11, 2005

### Gokul43201

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

10. Jan 11, 2005

### Curious3141

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

11. Jan 11, 2005

### Sombra

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

12. Jan 11, 2005

### recon

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

13. Jan 11, 2005

### Gokul43201

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

14. Jan 11, 2005

### Curious3141

5 is right.

15. Jan 11, 2005

### recon

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