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Finding Maximum and Minimum Values - Problem 1
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Here is another optimization problem and this one is very common you’ll often see it in either a pre-calculus or if not definitely a calculus course. Squares are to be cut from the corners of a 60 cm by 100 cm cardboard rectangle as shown in the picture to the right. The cardboard is then folded to make a box. What is the maximum volume of the box?

So I’ve taken this picture and drawn it on the board. This distance is 60 cm this is 100. Now I’m cutting squares out of each corner and each of the squares is going to be the same size. Let’s say that the side length of the square is x. Well then we can find the volume of the resulting box by noting that the volume is going to be the area of its base. This rectangle here will be the base of the box times the height so area of base times height.

The area of the base well it’s going to be length times width. And the length is 100 minus this x minus this x, 100 minus 2x. The width is 60 minus an x minus another x, 60 minus 2x. And the height after we fold up the sides, the height is going to be just x.

So this is our function, v equals 100 minus 2x, 60 minus 2x times x. Now the first thing I would do this is a polynomial function in fact it’s going to be a cubic. The first thing I’m going to do is plot the x intercepts now 1 is obvious 0,0 is an intercept that goes right here. What makes this 0? 30, so the 30 is going to be the intercept let’s make this point 30. And then 50 makes this 0 so 50 is going to be a x intercept.

And I want to flash this graph out a little bit it’s a cubic note that its lead term is going to be -2x times -2x times x or 4x cubed. So the right tail is going to go up and the left tail is going to go down, so it’s going to have this kind of shape but let’s figure out what it looks like in the middle.

So I have a table I’ll record some values. Let’s plot 10. If I plug in 10 I get 100 minus 20, 80 times 60 minus 20, 40, times 10. That’s 32 0-0-0. 32,000 let’s make this 32,000 right here and then let’s plug in 20 we have a 100 minus 40, 60, 60 minus 40, 20 and x is 20. So we have 12, 24 000. And 24,000 will go right here so this point.

So I have enough to draw a nice graph here, so something like this. Now I’m looking for the maximum volume of this box. This is the volume graph so I’m looking for the actual maximum value that it reaches. And unfortunately it's not until calculus that you actually learn an analytical way of doing this but we can use our calculator, our TI 84 to get the maximum value. So let me crack up my TI 84.

I want to enter my polynomial function in the y equals menu, I’ve already got it entered. 100 minus 2x 60 minus 2x and x. So let me graph that. Now the default in the TI 84 is to graph in the standard window minus 10 to 10 minus 10 to 10. Now obviously no part of our graph appears in the standard window but we remember that we graphed from 0 to 50. So let me change the window, x min I’m going to make 0 and x max I’m going to make 50, and let’s make the x scale 10.

Now x min let’s make that 0 and let’s think about what the y scale was on a hand drawn graph we’ve gone up to 32,000 let’s go up a little higher even, how about 48,000. And let’s go by increments of whatever 8,000 that’s fine. Alright let’s graph that.

Now you'll notice that we just kept the top piece of our volume function, now really this is the only piece that makes sense if you think about it. We want the volume to be positive. Now in order to find the location of this maximum value, your calculator actually has a command top do that.

Go into the calculate menu it's second trace and you have maximum right here option 4. So just hit number 4 we go back to graph mode it wants a left bound so pick a point on the left side of the maximum. And so you can cross all over or just kind of slow or you can just type in a number like I happen, I think 10 is going to work.

Right bound, we need to go to the right so let's pick 20. And I guess well something bigger than 10 looks like maybe 12. And the maximum occurs at 12.13, now that’s the x value. That’s the value of the size of the square cut out. This is the actual maximum volume 32,835 and that’s our answer.

So we need a calculator to get the answer to this optimization problem but next year in calculus you’ll learn analytical way to solve the problem.

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