# Finding Max/Min Values on Parametric Functions

1. Nov 12, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
Consider the function F(x,y) = 1 - x3 - y2 + x3y2. Consider the curve C given parametrically as x(t) = t1/3, y(t) = t1/2 for t ≥ 0. Determine the minimum and maximum of F(x,y) along the curve C.

3. The attempt at a solution
I think this is basically a max/min problem with a constraint function, so I will try to use Lagrange multipliers.

Fx = -3x2 + 3x2y2
Fy = -2y + x3*2y

Fx(t1/3,t1/2) = -3t2/3 + 3t5/3 = 1/(3t2/3) * λ
Fy(t1/3,t1/2) = -2t1/2 + 2t3/2 = 1/(2t1/2) * λ

9t4/9 * (-1 + t) = λ
4t1/4 * (-1 + t) = λ

9t4/9 = 4t1/4

t7/36 = 4/9
t = (4/9)36/7

And at this point I'm thinking there's no way this problem is this disgusting. Ideas on where I went wrong?

2. Nov 12, 2011

### Staff: Mentor

I don't get how you arrived at the last expression. The expression just before it can be factored to 3t2/3(-1 + t)
I'm not getting what you're doing here, either.

3. Nov 12, 2011

### TranscendArcu

I arrived at the last expression by setting

Fx(t1/3,t1/2) = -3t2/3 + 3t5/3

equal to the derivative of x(t), which I found to be 1/(3t2/3), and then multiplied by the constant λ, as required by the Lagrange multiplier method. Similarly, I let Fy(t1/3,t1/2) equal y'(t) multiplied by λ. By factoring, as you suggested, I can solve easily for λ, which is what you see above in,

9t4/9 * (-1 + t) = λ
4t1/4 * (-1 + t) = λ

However, I don't think this method is the proper way to go about answering the problem, since it gives a rather peculiar answer, namely t = (4/9)36/7

4. Nov 13, 2011

### HallsofIvy

Staff Emeritus
Personally, I wouldn't use Lagrange multipliers- it seems to me much easier just to substitute the values of x and y in terms of t. With $x= t^{1/3}$ and $y= t^{1/2}$, $F(x,y)= 1- x^3- y^2+ x^3y^2=1- t- t+ t^2= t^2- 2t+ 1= (t-1)^2$. Now, you don't even need to differentiate!