I am 99.9% certain no one has ever asked this question before. It's the optimization problem no one wants or needs but when you think about math as often as I do, this is the kind of thing you come up with. Now our family doesn't go out to eat much. When you have eight hungry kids you need a budget for such an occasion. One of our favorite places to go when the kids were younger was Arby's. For those that don't know Arby's is a fast food restaurant here in America that focuses on roast beef sandwiches. We all love the roast beef sandwiches mainly because the kids love to douse them in Arby's sauce. It's their house barbecue sauce. Once when we took the family out to eat at Arby's my son Spencer who's perhaps 15 went to the condiment counter to start loading up the little paper condiment cups with Arby sauce. And he loves Arby sauce, so he wanted to load a lot of sauce in his cup and started to spread the cup out by releasing some of the folds. This made the top of the cup a larger radius but it also lowered the height of the cup. So what did I do as a math nerd? I realized that there had to be an optimal angle for the sides to spread out so that the cup would hold the most sauce. If you opened it all the way out it would be flat and not hold much sauce and if you push the sides in it would decrease the amount of sauce the cup could hold. So somewhere in the middle there had to be a maximum volume. So I sat down, did the math so my son would know just how much to open up the cups to maximize the amount of his yummy Arby's sauce. So this is clearly math Overkill. My son could just get more cups if he wanted more sauce, but this is what you do as a math nerd. You find math problems and you just don't sleep well until you've taken a stab at solving them. So this week's version of math the world is a special Math Overkill Edition. The first thing I needed to do was to take some measurements and make some assumptions. I found a similar condiment cup online
and measured the height as 1 inch or 2.5 centimeters and the radius of the circular bottom is 1.5 centimeters or about 3
fifths of an inch. The sides are almost perpendicular to the base so I just took the 2.5 cm height to be the length of the side. I also assume that as we open up the cup that the rim stays a perfect circle, which isn't exact as you can see from this picture but it help simplify the problem for us. With the base and rim in perfect circles the shape is always a frustum of a cone, which is the part of a cone with the vertex cut off parallel to the base. There are a few ways to find the volume of this shape, someone did the math and derived the formula for a frustum. Which isn't too bad as long as you know the volume of a cone because a frustrum is a large cone minus a small cone. There's one trick at the end where it can be simplified by factoring the a difference of cubes so we can use the formula for a frustum volume. Or, since the frustum of a cone can be generated by rotating a region around an axis we could use a technique from calculus to calculate volumes of revolution. I did it both ways to check my answer, which is a move that would earn you a nerd sticker on your helmet of nerdiness. But seriously my students see how many mistakes I make when teaching, so I didn't trust myself until I had done it in two different ways and got the same answer. First we will use the frustum volume formula strategy, then later in the video set up the volume of Revolution strategy and let you finish it for another sticker on your nerd helmet. To use this frustrum formula we need three measurements, the height, the radius of the base, and the radius of the rim. Of these three quantities, only the radius of the base remains constant as the angle of the sides are changed up and down, so we need two functions. One that will give us the height if we know the angle of the side and another function that will give us the radius of the rim if we know the angle of the side. I'm going to define the angle Theta as this angle here, so Theta is zero when the sides are perpendicular to the base and 90° or Pi halves radians when the sides are parallel to the base. Drawing this triangle we can now use trig functions to get the height of the cup at a given Theta and the rim radius at a given Theta. The height is an adjacent leg to the angle in this right triangle so we can use the cosine function. This makes the height cosine of theta times the hypotenuse, which is 2.5 cm. The radius of the rim is the radius of the base plus this leg of the right triangle. This leg is opposite the angle so we can use the sin function and the hypotenuse to get that length sine of theta times 2.5. We can now input the radii and the height into the frustum volume formula to give us a formula for the volume of the cup given the angle of the sides. To find the maximum volume we have a couple of options. We can take the derivative with respect to Theta, set the derivative equal to zero and solve. Wolfram makes quick work of this. Or we could graph the function and have technology help us find the maximum. The graph shows us it's periodic and has many Maxima, but our context tells us we want the maximum between 0 and 90° or 0 and Pi halves radians. Either way we get a value of about 0.717 radians which is about 41°. Since this angle was the angle from vertical we want a cup that's a little more closed than open. In practice we can estimate the sides being about 45° but we
want the sides slightly higher than that. We've answered our initial question but we don't want to stop here when we have the chance to really Overkill this math problem and get a couple more cool nerd stickers on our golden rectangle helmet of nerdiness. Hey, if you're going to do Overkill math, you may as well go all the way right? We can solve this problem using an integral and a derivative if we find the volume using the volume of revolution technique from calculus. Let's set up the volume of revolution by putting the axes of the cup here so the origin is in the middle of the base of the cup. We need to find the equation of this line that goes from the outer edge of the base to the outer edge of the rim. let's start by finding the slope of that line which is dependent on the angle. We could do this from scratch with these points, or if we're clever instead of using a formula we can think of the meaning of the slope as the ratio of the change in the y direction, which is the height, and the change in the x direction, which is this horizontal leg of the triangle. And that ratio simplifies to just cotangent of theta. Now that we have the slope we can use the point slope formula with this point (1.5, 0) to get the equation of the line Y equals cotangent of theta times quantity (x - 1.5). The volume of revolution technique that we will use is called the disk method. If we chop up this volume horizontally so cuts are perpendicular to the Y axis in thickness of dy, then we can approximate the volume by adding up the volume of each of these cylinders. To find the volume of each cylinder, we need the height, which we call dy, and the area of the base. But for that we need the radius which changes depending on which value of y we are at. So we need to rewrite the equation of our line so the Y value is the input and the x value is the output. Now we can square the radius multiplied by pi, which is the area of the base of the cylinder, and multiply by dy to get the volume. If we continue with these products along the y-axis and add them up we get this integral. The bounds of this integral are all of the y-coordinates where we are accumulating volume so y equals 0 to y equal the height or y equals 2.5 times cosine of theta. Now for any angle Theta we have another way of finding the volume, by evaluating this integral. This isn't hard to do by hand since we're integrating with respect to Y so although that tangent of theta might look scary, with our nerd glasses on we see that it's actually a constant with respect to Y. But evaluating this integral isn't going to give us our answer. What we want is the maximum of this function. Now semi nerds might be thinking that we could take the derivative to find the maximum. So we have a derivative of an integral and that could save us a lot of work. The fundamental theorem of calculus does say say that those operations are opposites after all. But put on your nerd glasses and focus on this expression. Why can't we apply the fundamental theorem of calculus right now? It is because we are integrating with respect to Y and differentiating with respect to Theta. Since those aren't the same variables we can't apply the fundamental theorem of calculus. So we have to do the hard work of typing this into wolfram alpha and having it solve it for us. We get the same answer as before. So now that we've solved it two ways we can be more confident that our answer is correct. Now for those true math enthusiasts that have stuck around to the end of this video I have a homework problem for you. All along we've been assuming that the sauce goes just to the top of the rim. but those thick sauces can actually go much higher than the rim. What's the optimal angle if we fill the sauce to its max over the rim level, say 5 mm over the rim? Both the techniques we've shown in this video can help jump start you on this problem. Now the next time you're filling up your condiment cup at a fast food restaurant, maximize that sauce. Fold out that cup to a 41° angle and please meticulously relay all this information to whoever you are dining with. Thank you for watching! Please subscribe and share our videos. Be sure to follow math the world on Twitter, Instagram, and Facebook. Thank you so much for your support!
