Grades 11-12 Video Solutions 2010-11&12 Video Solutions 2010 problem24

Video Transcription

Question number 24. Consider all non-negative real numbers x and y whose sum is 2. What is the difference between the largest possible value and the  smallest possible value of the sum x cubed plus y cubed? So what we have to do is look at the function, call it A of x and y and that's the sum x cubed plus y  cubed and what we have to do is optimize. So find the largest and the smallest value of this expression. Now we can use the method of Lagrange multipliers and  use the function A directly here as a function of two variables and I will refer you to the suggested solutions for this method. I will solve this problem in  a different way using just the methods of standard first semester calculus and what I will do here is find the relationship between x and y so we can  solve here for y and obtain 2 minus x since x plus y is equal to 2 and we also have to note that both x and y are non-negative integers so the domain here  of A as a function of x now which is equal to x cubed plus and then we replace for y 2 minus x quantity cubed is the interval x greater than or equal  to 0 or an interval notation like this and so the max, min or extreme values occur at the end point so at x is equal to 0 and at the critical points so now I  will find the critical points let's save the information here about x is equal to 0 and for the critical points what we need to do is we need to differentiate  our function and set that equal to 0 so that happens if 0 is equal to and then we take the derivative here so 3x squared plus 3 quantity 2 minus x now to  the second power and there is a chain rule so you multiply by negative 1 here and so 0 has to be equal to 3x squared minus 3 times and then we have 4 minus 4x  plus x squared and that comes out to 3x squared on the right hand side minus 12 plus 12x minus 3x squared we see that the square terms cancel and so the  solution would be the equation is minus 12 plus 12x so the solution would be x is equal to 1 so that's a critical value and then we have our two values of x so  we have 1 x is equal to 0 y is equal to 2 and get then the sum x cubed plus y cubed is 8 when x is equal to 1 then y is equal to 1 and we get x cubed plus y  cubed is equal to 2 so obviously this is the maximum value this has to be the minimum value because we only have two critical points and then we subtract  them so the difference between them is 6 and we take that as our answer and that is choice B

Video Summary

The problem involves optimizing the expression \(x^3 + y^3\) for non-negative real numbers \(x\) and \(y\) such that \(x + y = 2\). By substituting \(y = 2 - x\) and analyzing the function over \(x\), critical points and endpoints can be identified to find extreme values. At \(x = 0\), \(y = 2\), resulting in the maximum value \(8\). At the critical point \(x = 1\), \(y = 1\), yielding the minimum value \(2\). Thus, the difference between the maximum and minimum values is \(6\), corresponding to choice B.

Keywords

optimization

expression

critical points

extreme values

non-negative