I am trying to do ques. 21 from stewart calculus. The question is given below.
a)
$\nabla f = \lambda \nabla g $
$\nabla g = (4x^3 – 3x^2) \hat i + (2y) \hat j$
$\nabla f = (1) \hat i $
System of Equations to find $(x,y)$ & $\lambda$
$1 = \lambda (4x^3 – 3x^2) $ — eqn. 1
$0 = \lambda (2y) $ — eqn. 2
$0 = y^2 + x^4 – x^3 $ — eqn. 3
$\lambda$ cannot be zero because of equation 1 even though equation 2 suggests.
therefore $y = 0$.
from eqn. 3 we can find $x$ for $y = 0$
$x = 0$ or $1$
If $x = 0$ equation 1 is Not satisfied. Therefore $x = 1$
For $x = 1$, $\lambda = 1$ from equation 1.
Finally, $x = 1,$ $y = 0$ and $\lambda = 1$ These should give the minimum/maximum point right?
I only have one point so how do I know if it is a max or min value? should I use the 2nd derivative test using hessian?
b) How to show minimum is at (0,0)?
I can easily show $\nabla f(0,0)$ $!= \lambda \nabla g(0,0) $ from above.
Please explain this first. I will try to do C by myself.
c)
Thanks.
Best Answer
The point of part (b) is to get an answer not using Lagrange multipliers.
$x^3-x^4=x^3(1-x)$ is equal to $y^2$ and so is non-negative. Therefore $0\le x\le1$ and the minimum is at $(0,0)$.
This also answers you query re. part (a). The point $(1,0)$ gives a maximum.