Minimize an cost function involving an integral

[Aerostd]

Homework Statement

The objective is to minimize

[itex]J(x) = \int y^2 (x+y)^2 dy, [/itex]

where x is the design variable.

The Attempt at a Solution

I first integrate over y to get
[itex] J(x) = \frac{x^2 y^3}{3} + \frac{y^5}{5} + \frac{2x y^4}{4}[/itex]
Now, I differentiate over x, and solve for the minimizer, the result is

[itex] x_{\textrm{min}} = \frac{-3y}{4}[/itex]

My question is the following: If I look at the cost function, I feel that the minimizer should be zero. Why am I getting [tex]\frac{-3y}{4}[/itex] as the minimzer?Background:
I heard somewhere that if you want to minimize

[itex] J(x) = \int f(y) f(x,y) dy [/itex],

where [itex]f(x)>0[/itex] and [itex]f(x,y)>0[/itex], then you can just obtain the minimizer by minimizing f(x,y). So I constructed the example above and it doesn't work. Can anyone point to a theorem that lists this kind of result?

Thanks.

 

[mighty2000]

Thank you for your post and for sharing your attempt at a solution. I can understand your confusion regarding the minimizer being [tex]\frac{-3y}{4}[/itex] instead of zero. However, the reason for this is because the cost function is a function of both x and y, and when you differentiate with respect to x, you are holding y constant. Therefore, the minimizer you found is the value of x that would minimize the cost function for a specific value of y. In order to find the overall minimizer, you would need to find the values of x and y that minimize the cost function simultaneously.

Regarding your background information, it is true that in some cases, minimizing a function of f(x,y) can give the same result as minimizing f(x) or f(y). However, this is not always the case, and it depends on the specific function and its properties. I am not aware of a specific theorem that lists this result, but it is a useful technique to keep in mind when solving optimization problems.

I hope this helps clarify your doubts. Best of luck with your research!