Minimum sum Find positive numbers x and y satisfying the equat...

Question

Minimum sum Find positive numbers x and y satisfying the equation xy = 12 such that the sum 2x + y is as small as possible.

Accepted Answer

Welcome back, everyone. Determine the minimum sum X + 4 Y for positive numbers X and Y satisfying the equation X Y equals 18. Were given 4 answer choices A says 22 2, B 42, C 62, and D 1222. So first of all, we're given our constraint X Y equals 18. And our objective function is X + or Y. We essentially want to minimize it, right? So what we're going to do is simply use the constraint function and express Y in terms of X. So basically Y is equal to 18 divided by X, and we're going to substitute this expression into our. Objective function. So we're going to get x + 4 multiplied by 18 divided by X, which gives us X + 72 divided by X. Let's call it P of X, our objective function. So what we want to do is simplify the derivative of PX. The derivative of X is equal to 1. And now the derivative of 1 divided by X is going to be equal to -1 divided by X2. So we get minus 72 divided by X2. And we're going to use the common denominator, which is X2. So, in the numerator, we're going to get X2 minus 72. Now what we want to do is simply set it equal to 0. First of all, To identify the critical points we're setting it to zero, but we also have to keep in mind that the derivative might not exist, but X cannot be equal to 0, so we're only solving this equation and not worrying about the existence of derivative because X. Is not equal to 0, right? Xy must be 18, so X cannot be equal to 0. This means that we're setting X2 minus 72 equals to 0. And essentially, we can solve this equation using the difference of squares. So it basically means that we're taking X2 minus square root of 722, which is equal to 0, and according to the difference of squares, this is X minus square root of 72 multiplied by X plus square root of 72, which is equal to 0. So this means that we have two solutions, X is either positive or negative, square root of 72, but X and Y, they are both positive. So we're taking X equals square root of 72. If we simplify, that's square root of 2 multiplied by 36, which is 6 square root of 2. What we want to do now is perform the 2nd derivative test. To prove that this is indeed a point of minimum, so we are going to identify the second derivative. P X. The derivative of 1 is 0 and the derivative of 1 divided by X2. Is equal to -1 divided by x cubed. We can essentially apply the power rule if we rewrite it as x to the power of -2, right? And because we're subtracting, well, what does that mean? minus gives us a positive value, so we're going to get 72 divided by X cubed. Now if we substitute. P double prime at point. 6 squad of 2, this is going to be positive, right, because we're dividing a positive number by a cube of a positive number and this means that the function is concave up at 62, and this indeed means that we have found a local minimum. So now we have proved that part and now we want to identify the minimum sum. So the minimum sum is our P of X. And specifically P at 0.6 square root of 2. Substituting, we're going to get X, which is 6 square of 2 + 72. Divided by 6 square root of 2. So we get 6 square root of 2, plus, now what is 72 divided by 6? Well, it's 12, so we get 12 divided by square root of 2. And if we rationalize the denominator, we're going to get 12 square root of 2. Divided by 2. And 12 divided by 2 simply 6, right? So we can just rewrite this term as 6 of 2. So our sum is 6 of 2 plus 6 words of T, which is 12 square root of T. Well then, we have found the minimum sum, and that corresponds to the answer to the 12 square root of 2. Thank you for watching.

Minimum sum Find positive numbers x and y satisfying the equation xy = 12 such that the sum 2x + y is as small as possible.

Key Concepts

Optimization

Constraints

Derivatives

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