Answer:
absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.
Step-by-step explanation:
We compute,
![$ f_x = 6x^2, f_y=4y^3 $](https://tex.z-dn.net/?f=%24%20f_x%20%3D%206x%5E2%2C%20f_y%3D4y%5E3%20%24)
Hence,
if and only if (x,y) = (0,0)
This is unique critical point of D. The boundary equation is given by
![$ x^2+y^2=1$](https://tex.z-dn.net/?f=%24%20x%5E2%2By%5E2%3D1%24)
Hence, the top half of the boundary is,
![$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}](https://tex.z-dn.net/?f=%24%20T%20%3D%20%5C%7B%20x%2C%20%5Csqrt%7B1-x%5E2%7D%20%3A%20-1%20%5Cleq%20x%20%5Cleq%201%5C%7D)
On T we have, ![$ f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1 \text{ for}\ -1 \leq x \leq 1$](https://tex.z-dn.net/?f=%24%20f%28x%2C%20%5Csqrt%7B1-x%5E2%7D%20%3D%202x%5E3%20%2B%281-x%5E2%29%5E2%20%3D%20x%5E4%20%2B2x%5E3-2x%5E2%2B1%20%20%5Ctext%7B%20for%7D%5C%20-1%20%5Cleq%20x%20%5Cleq%201%24)
We compute
![$ \frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$](https://tex.z-dn.net/?f=%24%20%5Cfrac%7Bd%7D%7Bdx%7D%28f%28x%2C%20%5Csqrt%7B1-x%5E2%7D%29%29%3D%204x%5E3%2B6x%5E2-4x%20%3D%202x%282x%5E2%2B3x-2%29%3D2x%282x-1%29%28x%2B2%29%3D0%24)
0 if and only if x=0, x= 1/2 or x = -2.
We disregard ![$ x = -2 \notin [-1,1]$](https://tex.z-dn.net/?f=%24%20x%20%3D%20-2%20%5Cnotin%20%5B-1%2C1%5D%24)
Hence, the critical points on T are (0,1) and ![$(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$](https://tex.z-dn.net/?f=%24%28%5Cfrac%7B1%7D%7B2%7D%2C%20%5Csqrt%7B1-%28%5Cfrac%7B1%7D%7B2%7D%29%5E2%7D%3D%5Cfrac%7B%5Csqrt3%7D%7B2%7D%29%24)
On the bottom half, B, we have
![$ f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$](https://tex.z-dn.net/?f=%24%20f%28x%2C%20%5Csqrt%7B1-x%5E2%7D%29%3D%20f%28x%2C-%5Csqrt%7B1-x%5E2%7D%29%24)
Therefore, the critical points on B are (0,-1) and
It remains to evaluate f(x, y) at the points
.
We should consider latter two points,
, since they are the boundary points for the T and also B. We compute ![$ f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2 $](https://tex.z-dn.net/?f=%24%20f%280%2C0%29%3D0%2C%20%5C%20%5Cf%280%20%5Cpm1%29%3D1%2C%20%5C%20%5C%20f%280%2C%20%5Cpm%20%5Csqrt3%2F2%29%3D9%2F16%2C%20%5C%20%5C%20f%281%2C0%20%29%3D%202%20%5Ctext%7B%20and%7D%5C%20%5C%20f%28-1%2C0%29%3D%20-2%20%24)
We conclude that the absolute maximum = f(1, 0) = 2
And the absolute minimum = f(−1, 0) = −2.