Optimization Problem

[JSmith]

Express 18 as a sum of two positive numbers whose product of the first and square of the second is as large as possible. Explain why this is the largest possible product.

First off, is this the proper answer? Secondly, I am unsure about how I should explain why this is the largest product... Is that sufficient?

 

[Deleted member 4993]

Express 18 as a sum of two positive numbers whose product of the first and square of the second is as large as possible. Explain why this is the largest possible product.

First off, is this the proper answer? Secondly, I am unsure about how I should explain why this is the largest product... Is that sufficient?

for a local maximum of P(y) we must have \(\displaystyle \dfrac{d^2P}{dy^2} < 0 \)

 

[HallsofIvy]

Your statement that "the restricted domain is \(\displaystyle 1\le x\le 17\)" is incorrect. It is 0< x< 18. (There is no requirement here that x be an integer.)
(Actually, it should be 0< y< 18 since the variable is y.)

You are correct that the derivative is \(\displaystyle 36y- 3y^2\) which is 0 for x= 0 and x= 12. One way of showing that x= 12 does actually give a maximum is, as Subhotash Kahn said, to look at the second derivative. P''= 36- 6y so P''(12)= 36- 72= -36< 0. Another way is to write P'= 3y(12- y) If y< 12, then 12-y is positive and so is P'. If y> 12, 12-y is negative so P' is negative. That means that the graph is increasing as we go up to 12 and then decreasing afterwards- a maximum at y= 12.