#
Optimization Using the Second Derivative Test - Problem 3
*
*2,441 views

The second derivative can also be used to find the dimensions of a poster with specified margin sizes to maximize the area. The area of the page is given, and the printed area is xy, where x and y are unknown. The margin sizes make the total side lengths of the poster x+margin and y+margin. So, the total area of the poster (which is given) would be A=xy, and the printed area of the poster is (x-2margin_{1})(y-2margin_{2}), where margin_{1} and margin_{2} are the given margin sizes. Substitute one value (x or y) with its relation from A=xy (y=A/x or x=A/y) to find that the printed area is (x-2margin_{1})(A/x-2margin_{2}) By finding the derivative, find the critical points of the function. Then, find the second derivative and evaluate it at the critical points. If the second derivative is positive at a point, that point is a maximum. Solve for y, and your answer will be the values of x and y.

Let's take a look at a tougher optimization problem. We're designing a poster to have an overall area of 300 square inches with 4 inch margins at the top, and bottom, and 3 inch margins on each side. Find the dimensions that will maximize the area of the printed material.

So here is the situation. We have this area of printed material in blue. This is the area I want to maximize. Then I have a page that's outlined in black here. That's the area that's fixed at 300 square inches.

So I'm told that this top margin, and the bottom margin are 4 inches. The side margins are 3. I'm told that the entire page's area is to be 300 square inches. So I need to introduce some variables. I think it's a good idea to make this x, and this y. So x times y, the area of the whole page, will be 300. That's my constraint.

Now, the dimensions that will maximize the area of the printed material. Let's try to express the area of the printed material in terms of the variables x and y. Well, if you look at what x is, x is this length plus 3 plus 3. So this length is x minus 6. If you add x minus 6 to 3 and 3, you get x. Then this length, the vertical length when you add 8 to it, you have to get y. So this length is y minus 8. Y minus 8, plus 4 plus 4 is y. So the area that I'm trying to maximize is x minus 6 times y minus 8.

Now this constraint is going to allow me to substitute for either y or x, whichever one I want to get rid of. Let me substitute for y. I'll make this y equals 300 over x. I can substitute in here. So I get area equals x minus 6 times 300 over x, that's my y value, minus 8. Let me multiply this through. 300 over x times x is 300, minus 8x, minus 1800 over x. And then plus 48. I have a total of 348 minus 8x minus 1800 over x. Now that's my area function.

I need to think about what the domain of this function is. So let's take a look at our picture again. Judging by this picture, what's the smallest that x can be? Because of the margins, I would say the smallest x can be is 6. When this dimension is actually 0, x has to at least be 6. Is there any maximum value? There maybe, but I think we can probably work with x greater than 6. So let me write that in.

Let's take this up here. I need to find the derivative, A'. The derivative of 348 is 0. The derivative of -8x, -8. The derivative of -1800 over x, that's the same as -1800 times x to the -1. It's going to be plus 1800x to the -2. So that's my derivative.

Now let me get a common denominator, and put these together, these two pieces. I have -8 plus 1800 over x². This can also be written with a denominator x². If I express it as -8x² over x². That will give me -8x² plus 1800 all over x².

Now critical points. x equals 0 is not a critical point. It's not in our domain. Our domain is x² greater than 6, so I can ignore that. Let me look at the numerator. This derivative will equal to 0 when 8x² equals 1800. Let me divide both sides by 8. 1800 over 8, so what's that? 900 over 4, 450 over 2, 225 nice. 225 a perfect square. So x equals 15. Not -15, it's not in the domain, but 15 is my only critical point.

So now I have to determine whether or not the second derivative is always positive or always negative. If it is, I can use the second derivative test. This is a pretty easy first derivative to differentiate again. So the second derivative is going to be -2 times 1800, -3600 x to the -3.

Now, keeping in mind that domain is x greater than 6, this part will always be positive, because this is the same as -3600 over x³. The x³ will always be bigger than 216, but this is going to be negative. So my second derivative is always negative. Now think about what that means? The graph will always be concave down. So my one and only critical point is going to be an absolute maximum. A has an absolute maximum at x equals 15.

Now let's go back and see what we were supposed to find. It says; find the dimensions that will maximize the area of the printed material. So x is 15. I just need to figure out what y is. Actually, I can look right at my constraint or this equation. X is 15 so y is going to be 300 over 15, that's 20. What are my units? Inches. So y equals 20 inches, x equals 15 inches. Those are the dimensions that will maximize the printed area.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete