QUESTION FOUR Find the minimum and the maximum values of $$ f(x, y)=8 x^{2}-2 y $$ subject to the constraint $$ x^{2}+y^{2}=1 $$ QUESTION FIVE Find the minimum and maximum values of $$ f(x, y, z)=y^{2}-10 z $$ subject to the constraint $$ x^{2}+y^{2}+z^{2}=36 $$

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To find the minimum and maximum values of the given functions subject to the specified constraints, we can use the method of Lagrange multipliers.

### QUESTION FOUR

We want to maximize and minimize the function:
$$
f(x, y) = 8x^2 - 2y
$$
subject to the constraint:
$$
g(x, y) = x^2 + y^2 - 1 = 0.
$$

Using Lagrange multipliers, we set up the equations:
$$
\nabla f = \lambda \nabla g.
$$

Calculating the gradients:
$$
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (16x, -2),
$$
$$
\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = (2x, 2y).
$$

Setting up the equations:
1. $$ 16x = \lambda (2x) $$
2. $$ -2 = \lambda (2y) $$
3. $$ x^2 + y^2 = 1 $$

From the first equation, if $$ x \neq 0 $$:
$$
\lambda = \frac{16x}{2x} = 8.
$$
From the second equation:
$$
-2 = \lambda (2y) \implies -2 = 8(2y) \implies -2 = 16y \implies y = -\frac{1}{8}.
$$

Substituting $$ y = -\frac{1}{8} $$ into the constraint:
$$
x^2 + \left(-\frac{1}{8}\right)^2 = 1 \implies x^2 + \frac{1}{64} = 1 \implies x^2 = 1 - \frac{1}{64} = \frac{63}{64} \implies x = \pm \sqrt{\frac{63}{64}} = \pm \frac{\sqrt{63}}{8}.
$$

Now we have two critical points:
1. $$ \left( \frac{\sqrt{63}}{8}, -\frac{1}{8} \right) $$
2. $$ \left( -\frac{\sqrt{63}}{8}, -\frac{1}{8} \right) $$

Calculating $$ f $$ at these points:
$$
f\left( \frac{\sqrt{63}}{8}, -\frac{1}{8} \right) = 8\left(\frac{63}{64}\right) + \frac{1}{4} = \frac{504}{64} + \frac{16}{64} = \frac{520}{64} = \frac{65}{8},
$$
$$
f\left( -\frac{\sqrt{63}}{8}, -\frac{1}{8} \right) = 8\left(\frac{63}{64}\right) + \frac{1}{4} = \frac{520}{64} = \frac{65}{8}.
$$

Next, we check the boundary points where $$ y $$ can take values $$ 1 $$ and $$ -1 $$:
1. For $$ y = 1 $$, $$ x = 0 $$: $$ f(0, 1) = 8(0) - 2(1) = -2 $$.
2. For $$ y = -1 $$, $$ x = 0 $$: $$ f(0, -1) = 8(0) - 2(-1) = 2 $$.

Thus, the minimum value is $$ -2 $$ and the maximum value is $$ \frac{65}{8} $$.

### QUESTION FIVE

We want to maximize and minimize the function:
$$
f(x, y, z) = y^2 - 10z
$$
subject to the constraint:
$$
g(x, y, z) = x^2 + y^2 + z^2 - 36 = 0.
$$

Using Lagrange multipliers, we set up the equations:
$$
\nabla f = \lambda \nabla g.
$$

Calculating the gradients:
$$
\nabla f = \left( 0, 2y, -10 \right),
$$
$$
\nabla g = \left( 2x, 2y, 2z \right).
$$

Setting up the equations:
1. $$ 0 = \lambda (2x) $$
2. $$ 2y = \lambda (2y) $$
3. $$ -10 = \lambda (2z) $$
4. $$ x^2 + y^2 + z^2 = 36 $$

From the first equation, if $$ x \neq 0 $$, then $$ \lambda = 0 $$. If $$ x = 0 $$, we can proceed with the other equations.

Assuming $$ x = 0 $$:
From the second equation:
$$
2y = \lambda (2y) \implies \lambda = 1 \text{ (if } y \neq 0\text{)}.
$$
From the third equation:
$$
-10 = \lambda (2z) \implies -10 = 2z \implies z = -5.
$$

Substituting $$ z = -5 $$ into the constraint:
$$
0 + y^2 + (-5)^2 = 36 \implies y^2 + 25 = 36 \implies y^2 = 11 \implies y = \pm \sqrt{11}.
$$

Now we have two critical points:
1. $$ (0, \sqrt{11}, -5) $$
2. $$ (0, -\sqrt{11}, -5) $$

Calculating $$ f $$ at these points:
$$
f(0, \sqrt{11}, -5) = (\sqrt{11})^2 - 10(-5) = 11 + 50 = 61,
$$
$$
f(0, -\sqrt{11}, -5) = (-\sqrt{11})^2 - 10(-5) = 11 + 50 = 61.
$$

Next, we check the case where $$ y = 0 $$:
From the constraint:
$$
x^2 + 0 + z^2 = 36 \implies x^2 + z^2 = 36.
$$
Then $$ f(x, 0, z) = -10z $$.

Using $$ z = \sqrt{36 - x^2} $$:
$$
f(x, 0, z) = -10\sqrt{36 - x^2}.
$$
To find the maximum and minimum, we can analyze the endpoints:
1. If $$ x = 0 $$, $$ z = 6 $$: $$ f(0, 0, 6) = -60 $$.
2. If $$ x = 6 $$, $$ z = 0 $$: $$ f(6, 0, 0) = 0 $$.

Thus, the minimum value is $$ -60 $$ and the maximum value is $$ 61 $$.

### Summary of Results
- **Question Four**: Minimum value is $$ -2 $$, Maximum value is $$ \frac{65}{8} $$.
- **Question Five**: Minimum value is $$ -60 $$, Maximum value is $$ 61 $$.
- For Question Four: Minimum value is -2, Maximum value is $$ \frac{65}{8} $$. 	- For Question Five: Minimum value is -60, Maximum value is 61.

QUESTION FOUR Find the minimum and the maximum values of \[ f(x, y)=8 x^{2}-2 y \] subject to the constraint \( x^{2}+y^{2}=1 \) QUESTION FIVE Find the minimum and maximum values of \[ f(x, y, z)=y^{2}-10 z \] subject to the constraint \( x^{2}+y^{2}+z^{2}=36 \)

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