12. find the maximum and the minimum values of the function subject to the given constraint or…

Explanation:

Step1: Set up the Lagrange function

For a function (f(x,y)) subject to a constraint (g(x,y)=c), the Lagrange function is (L(x,y,\lambda)=f(x,y)-\lambda(g(x,y) - c)). For part (a), (f(x,y)=x^{2}+y^{2}), (g(x,y)=x^{6}+y^{6}-1), so (L(x,y,\lambda)=x^{2}+y^{2}-\lambda(x^{6}+y^{6}-1)).

Step2: Take partial - derivatives

Calculate (\frac{\partial L}{\partial x}=2x - 6\lambda x^{5}=0), (\frac{\partial L}{\partial y}=2y - 6\lambda y^{5}=0), and (\frac{\partial L}{\partial\lambda}=-(x^{6}+y^{6}-1)=0). From (2x - 6\lambda x^{5}=0), we have (2x(1 - 3\lambda x^{4})=0), which gives (x = 0) or (\lambda=\frac{1}{3x^{4}}) (assuming (x\neq0)). Similarly from (2y - 6\lambda y^{5}=0), we have (y = 0) or (\lambda=\frac{1}{3y^{4}}) (assuming (y\neq0)).

Step3: Case 1: (x = 0)

If (x = 0), then from (x^{6}+y^{6}=1), we get (y=\pm1), and (f(0,\pm1)=1).

Step4: Case 2: (y = 0)

If (y = 0), then from (x^{6}+y^{6}=1), we get (x=\pm1), and (f(\pm1,0)=1).

Step5: Case 3: (x\neq0) and (y\neq0)

If (\lambda=\frac{1}{3x^{4}}) and (\lambda=\frac{1}{3y^{4}}), then (x^{4}=y^{4}), so (y=\pm x). Substituting (y = x) into (x^{6}+y^{6}=1), we have (2x^{6}=1), (x^{6}=\frac{1}{2}), (x=\pm\frac{1}{\sqrt[6]{2}}), and (f(\pm\frac{1}{\sqrt[6]{2}},\pm\frac{1}{\sqrt[6]{2}})=\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{2}}=\sqrt[3]{2}).

Step6: Determine maximum and minimum

The minimum value of (f(x,y)=x^{2}+y^{2}) subject to (x^{6}+y^{6}=1) is (\sqrt[3]{2}) and the maximum value is (1).

We can follow similar procedures for parts (b), (c), and (d) using the Lagrange - multiplier method. For part (b):

Step1: Set up the Lagrange function

(f(x,y)=6x - 8y), (g(x,y)=x^{2}+y^{2}-100), (L(x,y,\lambda)=6x - 8y-\lambda(x^{2}+y^{2}-100)).

Step2: Take partial - derivatives

(\frac{\partial L}{\partial x}=6 - 2\lambda x=0\Rightarrow\lambda=\frac{3}{x}(x\neq0)), (\frac{\partial L}{\partial y}=-8 - 2\lambda y=0\Rightarrow\lambda=-\frac{4}{y}(y\neq0)), (\frac{\partial L}{\partial\lambda}=-(x^{2}+y^{2}-100)=0).

Step3: Equate (\lambda) values

From (\frac{3}{x}=-\frac{4}{y}), we have (y=-\frac{4}{3}x).

Step4: Substitute into the constraint

Substitute (y = -\frac{4}{3}x) into (x^{2}+y^{2}=100), we get (x^{2}+\frac{16}{9}x^{2}=100), (\frac{9x^{2}+16x^{2}}{9}=100), (\frac{25x^{2}}{9}=100), (x^{2}=36), (x=\pm6). When (x = 6), (y=-8); when (x=-6), (y = 8). (f(6,-8)=6\times6-8\times(-8)=36 + 64 = 100), (f(-6,8)=6\times(-6)-8\times8=-36 - 64=-100). So the maximum is (100) and the minimum is (-100).

For part (c):

Step1: Set up the Lagrange function

(f(x,y,z)=x + 2y-2z), (g(x,y,z)=x^{2}+y^{2}+z^{2}-9), (L(x,y,z,\lambda)=x + 2y-2z-\lambda(x^{2}+y^{2}+z^{2}-9)).

Step2: Take partial - derivatives

(\frac{\partial L}{\partial x}=1 - 2\lambda x=0\Rightarrow\lambda=\frac{1}{2x}(x\neq0)), (\frac{\partial L}{\partial y}=2 - 2\lambda y=0\Rightarrow\lambda=\frac{1}{y}(y\neq0)), (\frac{\partial L}{\partial z}=-2 - 2\lambda z=0\Rightarrow\lambda=-\frac{1}{z}(z\neq0)), (\frac{\partial L}{\partial\lambda}=-(x^{2}+y^{2}+z^{2}-9)=0).

Step3: Equate (\lambda) values

From (\frac{1}{2x}=\frac{1}{y}), we have (y = 2x). From (\frac{1}{y}=-\frac{1}{z}), we have (z=-y=-2x).

Step4: Substitute into the constraint

Substitute (y = 2x) and (z=-2x) into (x^{2}+y^{2}+z^{2}=9), we get (x^{2}+4x^{2}+4x^{2}=9), (9x^{2}=9), (x=\pm1). When (x = 1), (y = 2), (z=-2), (f(1,2,-2)=1 + 2\times2-2\times(-2)=1 + 4 + 4 = 9). When (x=-1), (y=-2), (z = 2), (f(-1,-2,2)=-1+2\times(-2)-2\times2=-1 - 4 - 4=-9). So the maximum is (9) and the minimum is (-9).

For part (d):

Step1: Set up the Lagrange function

(f(x,y,z)=2x + 2y+z), (g(x,y,z)=4x^{2}+4y^{2}+z^{2}-9), (L(x,y,z,\lambda)=2x + 2y+z-\lambda(4x^{2}+4y^{2}+z^{2}-9)).

Step2: Take partial - derivatives

(\frac{\partial L}{\partial x}=2 - 8\lambda x=0\Rightarrow\lambda=\frac{1}{4x}(x\neq0)), (\frac{\partial L}{\partial y}=2 - 8\lambda y=0\Rightarrow\lambda=\frac{1}{4y}(y\neq0)), (\frac{\partial L}{\partial z}=1 - 2\lambda z=0\Rightarrow\lambda=\frac{1}{2z}(z\neq0)), (\frac{\partial L}{\partial\lambda}=-(4x^{2}+4y^{2}+z^{2}-9)=0).

Step3: Equate (\lambda) values

From (\frac{1}{4x}=\frac{1}{4y}), we have (x = y). From (\frac{1}{4x}=\frac{1}{2z}), we have (z = 2x).

Step4: Substitute into the constraint

Substitute (y = x) and (z = 2x) into (4x^{2}+4y^{2}+z^{2}=9), we get (4x^{2}+4x^{2}+4x^{2}=9), (12x^{2}=9), (x^{2}=\frac{3}{4}), (x=\pm\frac{\sqrt{3}}{2}). When (x=\frac{\sqrt{3}}{2}), (y=\frac{\sqrt{3}}{2}), (z=\sqrt{3}), (f(\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},\sqrt{3})=2\times\frac{\sqrt{3}}{2}+2\times\frac{\sqrt{3}}{2}+\sqrt{3}=3\sqrt{3}). When (x=-\frac{\sqrt{3}}{2}), (y=-\frac{\sqrt{3}}{2}), (z=-\sqrt{3}), (f(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2},-\sqrt{3})=-3\sqrt{3}). So the maximum is (3\sqrt{3}) and the minimum is (-3\sqrt{3}).

Answer:

(a) Maximum: (1), Minimum: (\sqrt[3]{2}) (b) Maximum: (100), Minimum: (-100) (c) Maximum: (9), Minimum: (-9) (d) Maximum: (3\sqrt{3}), Minimum: (-3\sqrt{3})