AP®︎/College Calculus AB
- Optimization: sum of squares
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- Optimization: box volume (Part 2)
- Optimization: profit
- Optimization: cost of materials
- Optimization: area of triangle & square (Part 1)
- Optimization: area of triangle & square (Part 2)
- Motion problems: finding the maximum acceleration
What is the minimum possible value of x^2+y^2 given that their product has to be fixed at xy = -16. Created by Sal Khan.
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- Well, but what is optimization ?(53 votes)
- Optimization just means finding the value that maximizes or minimizes something. In this case, we optimized (minimized) the sum of two numbers squared.(80 votes)
- What does calculus represent? What does it do for us?(3 votes)
- Our world is in constant motion, or change, so if you want to describe/explain/model/predict this world, the best tool to use is one that was developed to describe processes that change over time, namely the calculus; poetic license can share the emotive/descriptive aspect of change, but not tell us anything about how or why. Almost everything you see around you in the world today, internet, computers, highways, bridges, buildings, successful business, planes, trains and automobiles, medicines and medical devices, TVs, cell phones and even computer games could not be at the level of complexity they have now with out calculus.(53 votes)
- When you solve x^2=16 you said it's 4. But this could be -4 too, and so Y would be 4 instead of -4.(18 votes)
- As long as one integer is negative and one integer is positive in this case, then it doesn't matter which variable is negative or positive, it will still come out to be -16.(7 votes)
- The IWDDSP method to solving optimization problems:
1. Identify the quantity to be maximized/minimized and all given values.
2. Write an equation in terms of a single variable for the quantity to be maximized/minimized.
3. Determine the domain of the function.
4. Differentiate both sides of the equation with respect to the variable in the equation.
5. Set the derivative equal to zero and solve for the values that could maximize/minimize the desired quantity.
6. Plug the resulting values, along with the endpoints of the domain, into the equation to find the maximum/minimum quantity.(10 votes)
- At4:27: Wherefore do you take the second derivative of the original equation?(5 votes)
- The second derivative was to verify that the point x=4, would indeed create the smallest sum. And not, for instance, the largest sum. The first derivative only showed us that x=4 was a critical point, not which kind.(10 votes)
- When in real life would we use this?
and when would we use any other calculus stuff?(2 votes)
- If you wanna get into machine learning or tough scientific / engineering problems, this is a really important topic.
I'm 31 and learning this stuff - I need it for my job as a software engineer.(12 votes)
- If the graph is always concave upwards (since Y''(x) is always positive) how can there be two minimums (at 4 and -4)?(6 votes)
- Graphically, If x = +4 and -4 than function will have 2 points where derivative of f(x) = 0, means there will be two points where sope of f(x) = 0 . Where is the second point if Funcion is always concave upwards and doesn't have an inflection point ?(3 votes)
- The function has zero slope at x=4 and x=-4. It doesn't have an inflection point, but it has an asymptote at x=0 because one of the terms has x in the denominator and tends to infinity. It may appear to be a curve that's always concave up, but actually it's two separate curves that are concave up, with a discontinuity at x=0.(6 votes)
- Couldn't we just input a value in the first derivative that is close to x=4 from both the positive and negative sides. If the first derivatives gives a negative value for the side of x that is more negative, and the first derivative gives a value that is positive for the side of x that is more positive, then don't we know that x=4 is a minimum value? b/c the original function's slope is negative approaching from negative infinity and positive approaching from positive infinity?(4 votes)
- Correct, the first derivative test for local extrema is a valid alternative to the second derivative test for local extrema.(3 votes)
- Is the concept of maximisation and minimisation the same thing as convexity and concavity?(2 votes)
- Not exactly, but they are related.
Concavity is a description of how the slope of a function changes, and finding concavity involves looking at the second derivative of the parent function for zeros, which correspond to possible inflection points.
Optimization problems are problems of identifying certain extrema, and tend to involve not just finding them (which would be just looking at the first derivative of the parent function for zeros, which correspond to possible critical points/extrema) but also describing the parent function in the first place, determining it from a worded description of the problem. This is shown in the video here, where the word problem "minimize the sum of the squares of two numbers whose product is -16" must be translated into "minimize S(x), the single-variable function which represents the sum of the squares of two numbers whose product is -16".
The two concepts are related, in that the extrema found in optimization problems are usually at points that look like the tops of crests (maxima) or the bottoms of troughs (minima), and as a result the concavity around those extrema is very visually distinguishable.
(Note that when I say "possible" points, I mean that for it to be "actually" a critical or inflection point, the first or second derivative (respectively) would have to have an actual sign change in the neighborhood of the zero.)(7 votes)
We are being asked, what is the smallest-- this is a little typo here-- what is the smallest possible sum of squares of two numbers if their product is negative 16? So let's say that these two numbers are x and y. So how could we define the sum of the squares of the two numbers? So I'll just call that the sum of the squares, s for sum of the squares, and it would just be equal to x squared plus y squared. And this is what we want to minimize. We want to minimize s. Now, right now s is expressed as a function of x and y. We don't know how to minimize with respect to two variables, so we have to get this in terms of only one variable. And lucky for us, they give us another piece of information. Their product is negative 16. So x times y is equal to negative 16. So let's say we wanted this expression right over here only in terms of x. Well, then we can figure out what y is in terms of x and then substitute. So let's do that right over here. If we divide both sides by x, we get y is equal to negative 16 over x. And so let's replace our y in this expression with negative 16 over x. So then we would get our sum of squares as a function of x is going to be equal to x squared plus y squared. y is negative 16 over x. And then that's what we will now square. So this is equal to x squared plus, what is this? 256 over x squared. Or we could write that as 256x to the negative 2 power. That is the sum of our squares that we now want to minimize. Well, to minimize this, we would want to look at the critical points of this, which is where the derivative is either 0 or undefined, and see whether those critical points are possibly a minimum or a maximum point. They don't have to be, but those are the ones if we have a minimum or a maximum point, they're going to be one of the critical points. So let's take the derivative. So the derivative s prime-- let me do this in a different color-- s prime of x. I'll do it right over here, actually. The derivative s prime of x with respect to x is going to be equal to 2x times negative 2 times 2x plus 256 times negative 2. So that's minus 512x to the negative 3 power. Now, this is going to be undefined when x is equal to 0. But if x is equal to 0, then y is undefined. So this whole thing breaks down. So that isn't a useful critical point, x equals 0. So let's think about any other ones. Well, it's defined everywhere else. So let's think about where the derivative is equal to 0. So when does this thing equal 0? So when does 2x minus 512x to the negative 3 equal 0? Well, we can add 512x to the negative 3 to both sides. So you get 2x is equal to 512x to the negative third power. We can multiply both sides times x to the third power so all the x's go away on the right-hand side. So you get 2x to the fourth is equal to 512. We can divide both sides by 2, and you get x to the fourth power is equal to 256. And so what is the fourth root of 256? Well, we could take the square root of both sides just to help us here. So let's see. So it's going to be x squared is going to be equal to 256 is 16 squared. So this is 16. This is going to be x squared is equal to 16 or x is equal to 4. Now that's our only critical point we have, so that's probably the x value that minimizes our sum of squares right over here. But let's make sure it's a minimum value. And to do that, we can just do our second derivative test. So let's figure out. Let's take the second derivative s prime prime of x and figure out if we are concave upwards or downwards when x is equal to 4. So s prime prime of x is going to be equal to 2. And then we're going to have negative 3 times negative 512. So I'll just write that as plus 3 times 512. That's going to be 1,536. Is that right? Yeah, 3 times 500 is 1,500, 3 times 12 is 36, x to the negative 4 power. And this thing right over here is actually going to be positive for any x. x to the negative 4, even if the negative x value, that's going to be positive. Everything else is positive. This thing is always positive. So we are always in a concave upwards situation. Concave upwards means that our graph might look something like that. Actually, I don't want to draw the little squiggle. It might look something like that. And you see there's a reason why the second derivative implies concave upwards, a second derivative positive means that our derivative is constantly increasing. So the derivative is constantly increasing. It's negative, less negative, even less negative. Let me do it in a different color. You see it's negative, less negative, even less negative, 0, positive, more positive. So it's increasing over the entire place. So if you have a critical point where the derivative is equal to 0, so the slope is equal to 0, and it's concave upwards, you see pretty clearly that we have minimized the function. So what is y going to be equal to? We actually don't even have to figure out what y has to be equal to in order to minimize the sum of squares. We could just put it back into this. But just for fun, we see that y would be negative 16 over x. So y would be equal to negative 4. And we could just figure out now what our sum of squares is. Our minimum sum of squares is going to be equal to 4 squared, which is 16 plus negative 4 squared plus another 16, which is equal to 32. Now I know some of you might be thinking, hey, I could have done this without calculus. I could have just tried out numbers whose product is negative 16 and I probably would have tried out 4 and negative 4 in not too much time and then I would have been able to maybe figure out it's lower than if I did 2 and negative 8 or negative 2 and 8 or 1 and 16. And that's true, you probably would have been able to do that. But you still wouldn't have been able to feel good that that was a minimum value, because you wouldn't have tried out 4.01 or 4.0011. In fact, you couldn't have tried out all of the possible values. Remember, we didn't say that this is only integers. It just happened to be that our values just worked out to be integers in this situation. You can imagine what would happen if the problem wasn't if their product is negative 16, but what if their product is negative 17? Or what if the product is negative 16.5? Or what if their product was pi squared? Then you wouldn't be able to try everything else out and you would have to resort to doing what we did in this video.