Multivariable Calculus, circles, maximum and minimum

Question

Determine the minimum and maximum values of $f(x, y) = 2x^2 −3y^2 + 40y −6$ on the circle $x^2 + y^2 = 25$.

I don't understand what the question is asking me to do? What does it mean by "maximum and minimum values...on the circle..."?

Answer 1

The question asks the maximum and minimum values of the function $$f(x,y)=2x^2-3y^2+40y-6$$ subject to the constraint $x^2+y^2-25=0$

From the Lagrangian$:$ $$L(x,y,\lambda)=(2x^2-3y^2+40y-6)+\lambda(x^2+y^2-25)$$ Find all first order partial derivatives $$\frac{\partial}{\partial x}(2x^2-3y^2+40y-6)+\lambda(x^2+y^2-25)=2x(\lambda+2)$$ $$\frac{\partial}{\partial y}(2x^2-3y^2+40y-6)+\lambda(x^2+y^2-25)=2\lambda y-6y+40$$ $$\frac{\partial}{\partial x}(2x^2-3y^2+40y-6)+\lambda(x^2+y^2-25)=x^2+y^2-25$$ Now solve these three simulations liner equations that we got after taking partial derivatives. The solutions that will come are $$(x,y)=(3,4) ,(-3,4) ,(0,5) ,(0,-5)$$ $$\begin{align*} f(3,4)=124\\ f(-3,4)=124\\ f(0,5)=119\\ f(0,-5)=-281 \end{align*} $$ Hence the maximum value of the function subject to the given constraint is $\boxed{124}$ and minimum for the same is $\boxed{-281}$