# Who else but pf could help me w/this prob.

1. Mar 4, 2005

### karen03grae

So a friend of a friend calls me with this Cal. I. prob. I try to figure it out and get stuck. I have already told her that I don't know how to solve it. Now I am reallllyy curious on how to solve it. Here it goes:

You have a rectangular beam. Its cross-section has dimensions "l" and "w". The strength of the beam is directly proportional to w^2*l. Now if the beam is cut from a circular log with diameter of 3ft, what are the dimensions of the beam that will make it the strongest?

So I said s(w,l)= k*w^2*l where "s" is the strength. The width must be a little less than 3ft...but here I am just guessing. I know that l*w is proportional to pi*r^2, but I don't know where to go from there. help

2. Mar 5, 2005

### DoubleMike

That question was confunsingly worded, but I think it means the rectangular section you can cut our from the circle.

Hence, develop a formula for the dimensions of a circumscribed rectangle, then derive it.

3. Mar 5, 2005

### karen03grae

Yes! I can't relate the rectangle to the inscribed circle. How would one do that?

4. Mar 5, 2005

### HallsofIvy

Draw a picture. If a rectangle is inscribed in a circle, then its diagonal is a diameter of the circle. Now use the Pythagorean theorem.
Let R be the radius of the circle. If an inscribed rectangle has length and width, l and w, then $l^2+ w^2= (2R)^2$.

5. Mar 5, 2005

### karen03grae

I got w = 6/sqrt(5) and l= 3/sqrt(5). I said $$s(l,w) = w^2*l$$. Where "s" was the strength. I took the partial derivative with respect to both "l" and "w". The partial w.r.t. "l" was $$w^2$$. And the partial with respect to "w" was 2*l*w. I set both of these equal to zero. I did this to maximize the function. I don't think I was supposed to solve it this way because it is a problem from Cal. I. Anyway, I now have 2*l=w. I solved for "l" by the relationship that HallsofIvy told me to use. So $$l=sqrt(9-w^2)$$ gets substituted into l= w/2 and that's how I got w= 6/sqrt(5). It sounds reasonable, but I would like to know if I did it correctly.

6. Mar 5, 2005

### DoubleMike

You could've used the fact that w^2 = (9-l^2)

and just subed that directly into the equation.

7. Mar 5, 2005

### so-crates

No, your answer is wrong :-) Follow double Mikes suggestion. There is no need for implcit differentiation

8. Apr 1, 2005

### karen03grae

Never did figure out this one.

9. Apr 1, 2005

### HallsofIvy

You are given that the strength is s= kw l2 and you know that
w2+ l2= 4R2 so l2= 4R2- w2 and s= 4kR2w- kw3.

Differentiating with respect to w, s'= 4kR2- 3kw2= 0 at a critical value. That is, w2= (4/3)R2 so that $$w= \sqrt{\frac{4}{3}}R$$ and then l2= 4R2- (4/3)R2= (8/3)R2 and $$l= 2\sqrt{\frac{2}{3}}R$$.

l, you might notice, is not "w/2", it's w= l/2.

10. Apr 1, 2005

### whozum

Thats a tough problem for calc 1. Perhaps latex is just making it look really complicated.

11. Apr 2, 2005

### karen03grae

Thanx. Can rest easier now.