The Lagrangian is
![L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)](https://tex.z-dn.net/?f=L%28x%2Cy%2C%5Clambda%29%3Dx%2B4y%2B%5Clambda%28x%5E2%2By%5E2-9%29)
with critical points where the partial derivatives vanish.
![L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}](https://tex.z-dn.net/?f=L_x%3D1%2B2%5Clambda%20x%3D0%5Cimplies%20x%3D-%5Cdfrac1%7B2%5Clambda%7D)
![L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda](https://tex.z-dn.net/?f=L_y%3D4%2B2%5Clambda%20y%3D0%5Cimplies%20y%3D-%5Cdfrac2%5Clambda)
![L_\lambda=x^2+y^2-9=0](https://tex.z-dn.net/?f=L_%5Clambda%3Dx%5E2%2By%5E2-9%3D0)
Substitute
into the last equation and solve for
:
![\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6](https://tex.z-dn.net/?f=%5Cleft%28-%5Cdfrac1%7B2%5Clambda%7D%5Cright%29%5E2%2B%5Cleft%28-%5Cdfrac2%5Clambda%5Cright%29%5E2%3D9%5Cimplies%5Clambda%3D%5Cpm%5Cdfrac%7B%5Csqrt%7B17%7D%7D6)
Then we get two critical points,
![(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)](https://tex.z-dn.net/?f=%28x%2Cy%29%3D%5Cleft%28-%5Cdfrac3%7B%5Csqrt%7B17%7D%7D%2C-%5Cdfrac%7B12%7D%7B%5Csqrt%7B17%7D%7D%5Cright%29%5Ctext%7B%20and%20%7D%28x%2Cy%29%3D%5Cleft%28%5Cdfrac3%7B%5Csqrt%7B17%7D%7D%2C%5Cdfrac%7B12%7D%7B%5Csqrt%7B17%7D%7D%5Cright%29)
We get an absolute maximum of
at the second point, and an absolute minimum of
at the first point.
You can use it to find the area of something
Answer:
8.06
Step-by-step explanation:
Hope it helps
Answer:
A 7,4.5
Step-by-step explanation:
Answer:Roots: 3;2;-1
Step-by-step explanation: i use horner's scheme for approximating the roots of polynomials