- #36
MarkFL
Gold Member
MHB
- 13,288
- 12
Yes, we actually gain a critical point to check that we did not get from the parametrization of the boundary...but does this affect the absolute extrema values?
I do not know? I Should check all critical point so yes?MarkFL said:Yes, we actually gain a critical point to check that we did not get from the parametrization of the boundary...but does this affect the absolute extrema values?
The crit point when we subsitate x. But the point is you can't divide by zero so when we did calculate that 3,-3 we say that the bottom of division is zero..?MarkFL said:As a recap, what are all of the critical points you have collected so far, and what is the function's value at these points?
Yeah my bad I forgot that we did subsitute for y. One question what did you mean with interior (1,2)MarkFL said:The derivative is undefined for $x^2=3^2$, but the original function is not. Recall that critical values come from places where the derivative is zero or undefined.
You should have the critical point in the interior (1,2), and the points on the boundary:
\(\displaystyle \left(\pm\frac{3}{\sqrt{5}},\pm\frac{6}{\sqrt{5}} \right),\,(\pm3,0)\)
So, you want to evaluate the original function at these 5 points...
Petrus said:...One question what did you mean with interior (1,2)
Petrus said:Ok for Lama way.
We got our \(\displaystyle g(x,y)=x^2+y^2-9\)
so..
\(\displaystyle f_x(x,y)=\lambda*g_x(x,y)\)
\(\displaystyle f_y(x,y)-=\lambda*g_y(x,y)\)
And our...
\(\displaystyle g_x(x,y)=2x\)
\(\displaystyle g_y(x,y)=2y\)
so we now got..
\(\displaystyle f_x(x,y)=\lambda*2x\)
\(\displaystyle f_y(x,y)-=\lambda*2y\)
Now I don't know to do but I guess ima subsitute again \(\displaystyle y=\sqrt{9-x^2}\) and then equal each to same so I got
\(\displaystyle 2x=2\sqrt(9-x^2}*\)
I am doing correct?
I get x=9 and x=7MarkFL said:Recall this is the critical point you found in the interior of the circle by equating the first partials to zero.
Using Lagrange multipliers, you could state $f(x,y)=17-2x-4y$ as the objective function and $g(x,y)=x^2+y^2-9=0$ as the constraint. This gives you the system:
\(\displaystyle -1=\lambda x\)
\(\displaystyle -2=\lambda y\)
What does this imply about the relationship between $x$ and $y$? Solve both for $\lambda$, then equate and simplify. Once you find this, substitute for $y$ in the constraint, and solve for $x$. This will give you two critical points on the boundary.
\(\displaystyle \lambda=-\frac{1}{x}\) and \(\displaystyle -\lambda=\frac{2}{y}\)MarkFL said:Recall this is the critical point you found in the interior of the circle by equating the first partials to zero.
Using Lagrange multipliers, you could state $f(x,y)=17-2x-4y$ as the objective function and $g(x,y)=x^2+y^2-9=0$ as the constraint. This gives you the system:
\(\displaystyle -1=\lambda x\)
\(\displaystyle -2=\lambda y\)
What does this imply about the relationship between $x$ and $y$? Solve both for $\lambda$, then equate and simplify. Once you find this, substitute for $y$ in the constraint, and solve for $x$. This will give you two critical points on the boundary.