## Difficulty with optimization problems

I'm going through a calculus textbook in an attempt to learn it myself. So far so good, but I've been stuck on optimization problems.
I understand the concept. The maxima and minima of a function can be found by looking at where its derivative = 0.
I also see that a function that has no upper or lower bound can still have local maxima and minima.
I went through the problems in max/min section of my textbook just fine.

But I have run into trouble in the optimization section.

In these problems, one is asked to maximize or minimize some value (say the volume of a box, or the area of a window, or whatever).
First, one writes a function, almost always of more than one variable.
Secondly, using other information, one finds an expression to substitute into the function and write it in terms of a single variable.
Third, one differentiates the function, looks for the zeroes, and then evaluates the initial function at these values etc.

My problem is in step two. I can't find ways of writing functions in terms of a single variable.
Here's one that gives me trouble:

A right circular cylinder is inscribed in a sphere of radius R. What is the maximum surface area of such a cylinder?

So, the cylinder is inside the sphere, its edges touching the sphere.
We can call its radius c, and its height h.

Its area is:

$$A = 2 \pi c^2 + 2 \pi c h$$

Now we have to differentiate and find A' . But how do we find a substitute for h?

From the cross section of the cylinder, we can see that the cylinder's radius and height/2 form a right triangle with the radius R of the sphere. So:

$$R^2 = c^2 + (h/2)^2$$ Right?

solving for h, we have:

$$h = 2 \sqrt {R^2 - c^2}$$

We then substitute h into the area function, and differentiate. I get nothing like what my textbook gives, which is $$\pi R^2(1+ \sqrt{5})$$

I don't see how they got this.
 nothing?

Mentor
 Quote by hexag1 I'm going through a calculus textbook in an attempt to learn it myself. So far so good, but I've been stuck on optimization problems. I understand the concept. The maxima and minima of a function can be found by looking at where its derivative = 0. I also see that a function that has no upper or lower bound can still have local maxima and minima. I went through the problems in max/min section of my textbook just fine. But I have run into trouble in the optimization section. In these problems, one is asked to maximize or minimize some value (say the volume of a box, or the area of a window, or whatever). First, one writes a function, almost always of more than one variable. Secondly, using other information, one finds an expression to substitute into the function and write it in terms of a single variable. Third, one differentiates the function, looks for the zeroes, and then evaluates the initial function at these values etc. My problem is in step two. I can't find ways of writing functions in terms of a single variable. Here's one that gives me trouble: A right circular cylinder is inscribed in a sphere of radius R. What is the maximum surface area of such a cylinder?
R should be considered a constant.
 Quote by hexag1 So, the cylinder is inside the sphere, its edges touching the sphere. We can call its radius c, and its height h.
I would call its radius r (which is different from R, the radius of the sphere).
You can simplify things by looking at a cross-section, so that you have a rectangle inscribed within a circle of radius R. The radius of the cylinder is one-half the width of the rectangle, and the height of the cylinder is the same as the height of the rectangle, or h.

The rectangle intersects the circle at the points (r, h/2), (r, -h/2), (-r, h/2), (-r, -h/2). I'm assuming the circle is centered at (0, 0). Since you know the equation of the circle (right?), you can get a relationship between r and h to eliminate one variable.
 Quote by hexag1 Its area is: $$A = 2 \pi c^2 + 2 \pi c h$$ Now we have to differentiate and find A' . But how do we find a substitute for h? From the cross section of the cylinder, we can see that the cylinder's radius and height/2 form a right triangle with the radius R of the sphere. So: $$R^2 = c^2 + (h/2)^2$$ Right? solving for h, we have: $$h = 2 \sqrt {R^2 - c^2}$$ We then substitute h into the area function, and differentiate. I get nothing like what my textbook gives, which is $$\pi R^2(1+ \sqrt{5})$$ I don't see how they got this.

 Tags calculus, cylinder, differentiation, optimization, sphere