Question

1. find the maximum and the minimum values of the function subject to the given constraint or constraints.

(a) (f(x,y)=x^{2}+y^{2}) subject to (g(x,y)=x^{6}+y^{6}=1).
(b) (f(x,y)=6x - 8y) subject to (g(x,y)=x^{2}+y^{2}=100).
(c) (f(x,y,z)=x + 2y-2z) subject to (g(x,y,z)=x^{2}+y^{2}+z^{2}=9).
(d) (f(x,y,z)=2x + 2y+z) subject to (g(x,y,z)=4x^{2}+4y^{2}+z^{2}=9) and (h(x,y,z)=x^{2}+y^{2}+4z^{2}=9)

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 \(x eq0\)). Similarly from \(2y - 6\lambda y^{5}=0\), we have \(y = 0\) or \(\lambda=\frac{1}{3y^{4}}\) (assuming \(y eq0\)).

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

eq0\) and \(y eq0\)
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

\(\frac{\partial L}{\partial x}=6 - 2\lambda x=0\Rightarrow\lambda=\frac{3}{x}(x eq0)\), \(\frac{\partial L}{\partial y}=-8 - 2\lambda y=0\Rightarrow\lambda=-\frac{4}{y}(y eq0)\), \(\frac{\partial L}{\partial\lambda}=-(x^{2}+y^{2}-100)=0\).

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

\(\frac{\partial L}{\partial x}=1 - 2\lambda x=0\Rightarrow\lambda=\frac{1}{2x}(x eq0)\), \(\frac{\partial L}{\partial y}=2 - 2\lambda y=0\Rightarrow\lambda=\frac{1}{y}(y eq0)\), \(\frac{\partial L}{\partial z}=-2 - 2\lambda z=0\Rightarrow\lambda=-\frac{1}{z}(z eq0)\), \(\frac{\partial L}{\partial\lambda}=-(x^{2}+y^{2}+z^{2}-9)=0\).

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

\(\frac{\partial L}{\partial x}=2 - 8\lambda x=0\Rightarrow\lambda=\frac{1}{4x}(x eq0)\), \(\frac{\partial L}{\partial y}=2 - 8\lambda y=0\Rightarrow\lambda=\frac{1}{4y}(y eq0)\), \(\frac{\partial L}{\partial z}=1 - 2\lambda z=0\Rightarrow\lambda=\frac{1}{2z}(z eq0)\), \(\frac{\partial L}{\partial\lambda}=-(4x^{2}+4y^{2}+z^{2}-9)=0\).

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: \(-3\sqrt{3}\)