Optimisation Problems

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Think of all the rectangles whose perimeter is forty. Take a long thin one, nineteen by one. Its area is just nineteen. Now fifteen by five: area seventy five. Much better! Same perimeter, very different areas, right? So which rectangle gives the largest area of all? Maybe you have a guess. But can you prove it? That is what optimisation means: making a quantity as large, or as small, as possible, using derivatives. These problems come up often in JEE papers, so let's begin.

0:37The method: three steps

So, here is the method, in three steps. Step one: name the quantity to be optimised, the thing you want largest or smallest, like an area or a volume. And name the variables it depends on. Okay? Now, very often the quantity has two variables. The area of a rectangle, for example, is length times breadth. But our derivative works on a function of one variable. So, what do we do? Use the constraint. Yes, exactly! The question always gives a condition that ties the variables together, like a fixed perimeter. Use it to write one variable in terms of the other. Now the quantity is a function of just one variable. And note its allowed interval. Remember, a length can never be negative. Many students skip this, and it costs them. Right? Step three is pure calculus. Find the critical points, test them, and check the endpoints. Three steps, every time. Now let us see what testing and checking really mean. Clear?

1:40Step three: test, and check the ends

Now, a quick reminder. f prime of x, the f with a little dash, is the derivative. So a critical point c is a point where f prime of c is zero, or where f prime of c does not exist. Got it? First rule: keep only the critical points inside the allowed interval. Suppose a critical point gives a negative length. It means nothing for our problem, so drop it. With me? Now, say f prime of c is zero, so the tangent there is flat. Tell me, is c a maximum or a minimum? We cannot say yet! It could be either, or even neither. So, we test it. f double prime, with two little dashes, is the second derivative. With f prime of c equal to zero: if f double prime of c is less than zero, c is a local maximum. If it is greater than zero, c is a local minimum. And if it is zero, the test fails, so we cannot use it there. Okay? Next, the endpoints. If f is continuous on the closed interval a comma b, it surely reaches a largest and a smallest value there. The square brackets mean both ends are included. To find them, compare the values of f at the critical points inside, and at the two ends, a and b. Clear? Now, a trap. Suppose you have found a local maximum. That is surely the largest value, right? No, no, careful! Look at this one, from the previous concept on absolute extrema. f of x equals two x cubed, minus fifteen x squared, plus thirty six x, plus one, on the closed interval one comma five. It has a local maximum at x equals two, where f is twenty nine. But at the end point five, f is fifty six. Much bigger! Local means largest only nearby. So always check the ends. Got it?

3:38Example 1: largest area rectangle with perimeter 40

Now, remember our puzzle from the start? Let's finish it. Of all rectangles with perimeter forty, which has the largest area? Use the three steps. Pause the video, and try it first. Done? Okay, let's check. Step one: the quantity is the area, and the sides are x and y. Step two: the perimeter, two x plus two y, equals forty. The double arrow means implies, so y equals twenty minus x. Both sides must be positive, so x lies between zero and twenty. Right? So the area, capital A, equals x times the bracket, twenty minus x. One variable, just as we wanted. Got it? Step three: differentiate. Where is A prime zero? Yes, at x equals ten! The derivative, A prime, is twenty minus two x, and that is zero when x is ten. And A double prime is minus two, less than zero. So x equals ten is a maximum, and it is the only critical point, so it gives the largest area. Okay? Then y is twenty minus ten, also ten. So the winner is a square, ten by ten. The cross between the tens is read as by. And its area is one hundred. Puzzle solved! Nineteen, then seventy five, and now one hundred, the largest area any rectangle with perimeter forty can have. With me?

5:10Example 2: largest open box from an 18 by 18 sheet

Now a harder one, a classic. Equal squares are cut from the corners of an eighteen by eighteen sheet, and the sides are folded up to make an open box. Find the largest volume. Pause and try. Hint: call the side of each cut square x. Done? Okay, let's check. Let each cut square have side x. Each edge of the sheet loses x at both ends, so the base is eighteen minus two x, by eighteen minus two x. Fold the flaps up, and the height of the box is x. Right? Now step two, the allowed interval. What values can x take? Yes! x must be positive, and the base side, eighteen minus two x, must be positive too. So x is less than nine. So the volume, capital V, equals x times eighteen minus two x, the whole squared. The little two up there means squared. And x lies between zero and nine. Clear? Step three. V prime, the derivative, factorises as eighteen minus two x, times eighteen minus six x. It is zero at x equals three, and at x equals nine. Okay? So we have two candidates, three and nine. Both give a largest box, right? No, no, careful! This is the trap. At x equals nine, the base side is eighteen minus eighteen, which is zero. No base, no box, so the volume is zero. Nine is an end of the interval, not a point inside it. And at the other end, x equals zero, nothing is cut, so again the volume is zero. Got it? So x equals three is the only critical point inside. The volume is zero at both ends and positive in between, so three must give the largest. The base side is eighteen minus six, that is twelve, and twelve squared is one hundred and forty four. So V equals three times one hundred and forty four: four hundred and thirty two. See? The three steps did all the work. With me?

7:21Quick revision: constraint, critical points, endpoints

So, quick revision. Use the constraint to get one quantity in one variable, with its allowed interval. Keep only the critical points inside the interval, and test them. And always compare with the endpoints before you answer. Now go and try the practice problems on this page. You will enjoy them, I promise. All the best!