Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Three positive numbers, x, y and z, with x plus two y plus three z equal to twelve. How big can the product x times y squared times z cubed get? Try seven, one and one. Seven plus two plus three is twelve, and the product is just seven. Now try four, one and two. Four plus two plus six is twelve, and the product is four times one times eight: thirty two. Better! So, can it reach one hundred? Hmm, hard to be sure, right? Today we settle it in a few lines. This is Set three of our mixed problems on sequences and series, and questions like this come up often in JEE Main. So, let's begin.

0:47Set 3: what is inside

So, here is what Set three holds. Eight solved problems: the A P and the G P, the means, and sums to infinity. All mixed up, just like the real exam. Okay? Today we build one tool, the chain of means, then solve two problems from the set: an easy one first, then a harder one. With me?

1:09The chain of means

Now, the tool. Take n numbers, a one, a two, up to a n: the letter a with a small number below it, and the dots mean and so on. Their arithmetic mean, the A M, is their sum over n. Okay? Next comes the G M, the geometric mean. Multiply all n numbers, and take the n th root: that is the root sign with a small n sitting on it. The raised dots in the middle just mean keep multiplying. Clear? And the third one is the H M, the harmonic mean. Flip every number: one over a one, one over a two, and so on. Add these reciprocals, and take n over that sum. So the H M is built from reciprocals. Got it? Now tell me. For positive numbers, which of these three means is the largest? The A M, yes! The A M is greater than or equal to the G M, and the G M is greater than or equal to the H M, the smallest. That sign, greater than with a line below it, is read greater than or equal to. But the chain needs positive numbers, so always check that first. Right? And all three become exactly equal when all the numbers are equal, and only then. This equality case tells us where a least or a greatest value is actually reached. Okay?

2:35Least sum, greatest product

So, how do we use this chain? For the least value of a sum, make the product of the terms constant. Then the G M is fixed, and the sum can never fall below n times it. With me? Now flip it. We want the greatest value of a product. What should be constant this time? The sum. Yes, exactly! Fixed sum means a fixed A M, and the G M can never go above it. So the product has a ceiling. Clear? And if the variables in the product carry powers? Split the sum so that each variable appears as many times as its power: two copies for a square, three for a cube. Right? And two checks, every single time. All the terms must be positive, and equality must really be possible. Miss either one, and the answer can go wrong. Got it?

3:25Example 1: least value of 1/a + 1/b + 1/c if a + b + c = 3

Now, the first problem from the set, an easy one. Small a, b and c are greater than zero, so positive, and a plus b plus c equals three. Find the least value of one over a, plus one over b, plus one over c. Pause the video, and try it first. Done? Okay, let's check. We want the least value of a sum. So the A M and G M on the three reciprocals should do it, right? No, no, wait! Look at their product. One over a, times one over b, times one over c; the dot on screen means times. That is one over a b c, and only the sum of a, b and c is fixed, not their product. So that bound keeps moving, and it needs one more step to finish. There is a quicker way. Okay? So, which mean is built from reciprocals? The H M, yes, exactly! For the positive numbers a, b and c, the A M is at least the H M, so a plus b plus c, over three, is greater than or equal to three over the sum of the reciprocals. Clear? Now use the given sum. Since a plus b plus c is three, the left side is three over three, just one. Right? Multiply both sides by the sum of the reciprocals; it is positive, so the sign stays. The double arrow means, so: the sum of the reciprocals is at least three. With me? But is three really reached, or only a floor we never touch? Equality needs a, b and c all equal, and they add to three, so each one is one. Then one plus one plus one is three. So the least value is three. Perfect!

5:04Example 2: greatest xy^2z^3 when x + 2y + 3z = 12

Now, back to our puzzle from the start. x, y and z are positive, and x plus two y plus three z equals twelve. Find the greatest value of x y squared z cubed: the small two and three up top are the powers. Pause the video, and try it. Hint: the sum is fixed. Done? Okay, let's check. The sum is fixed, good. So use the A M and G M on the three terms x, two y and three z, right? Oh no, careful! Their product is x times two y times three z: six x y z. But we want x y squared z cubed. Wrong powers, so that bound is about the wrong product. That is the trap. Got it? So, y carries the power two, and z the power three. How do we get y twice and z three times in the sum? Split them. Yes! Two y is y plus y, and three z is z plus z plus z. So the sum is now six positive terms, still adding to twelve. Okay? Now the chain. The A M of the six terms is twelve over six, which is two, and it is at least the G M: the sixth root of their product, with a small six on the root sign. And x times y times y times z times z times z is exactly x y squared z cubed. The powers match! Right? Raise both sides to the power six. So x y squared z cubed is less than or equal to two to the power six; that sign is less than with a line below it. And two to the power six is sixty four. Clear? But is sixty four really reached? Equality needs all six terms equal, and each one must be the A M, two. So x, y and z are all two. Check: two plus four plus six is twelve. With me? So the greatest value is sixty four. Puzzle solved! Our hook found seven, then thirty two, but the true top is sixty four, at two, two and two. So one hundred is impossible. That's it!

7:11Quick revision: AM, GM, HM chain and splitting terms

So, quick revision. One: for positive numbers, the A M is at least the G M, which is at least the H M, with equality only when all the numbers are equal. Two: for a least sum, keep the product constant; for a greatest product, keep the sum constant. Three: reciprocals point to the A M and the H M, and powers mean you split the terms to match. Now go and try the practice problems on this page. You are ready, I promise. All the best!