A. Use Lagrange Multiplier (LM) to find points on the surface z=x2+y2 closest to the point (0,0,2). b. Consider the two surfaces given by: S1 = x2+y2+z2−4x+1=0 and S2 = y2−x+1. Find the maximum/minimum distance for points in both S1 and S2. (Hint: Use LM)

We want to minimize the distance between a point $(x,y,z)$ on the surface and $(0,0,2)$. It's easier to minimize the square of the distance:

The constraint is that the point $(x,y,z)$ lies on the surface $z = x^2 + y^2$, which can be written as:

We need to solve $\nabla f = \lambda \nabla g$.

The system of equations is:

1. $2x = \lambda(2x)$
2. $2y = \lambda(2y)$
3. $2(z-2) = \lambda(-1)$
4. $x^2 + y^2 - z = 0$ (the constraint)

From (1), $2x(1-\lambda) = 0$, which implies $x=0$ or $\lambda=1$.
From (2), $2y(1-\lambda) = 0$, which implies $y=0$ or $\lambda=1$.

• Case 1: $x=0$ and $y=0$ Substitute into (4): $0^2 + 0^2 - z = 0 \implies z=0$. So, the point is $(0,0,0)$. Check with (3): $2(0-2) = -\lambda \implies -4 = -\lambda \implies \lambda=4$. This is consistent.

• Case 2: $\lambda=1$ Substitute into (3): $2(z-2) = -1 \implies 2z - 4 = -1 \implies 2z = 3 \implies z = \frac{3}{2}$. Substitute $z = \frac{3}{2}$ into (4): $x^2 + y^2 - \frac{3}{2} = 0 \implies x^2 + y^2 = \frac{3}{2}$. This gives a set of points $(x,y,z)$ where $x^2+y^2 = \frac{3}{2}$ and $z=\frac{3}{2}$.

• For $(0,0,0)$: $f(0,0,0) = 0^2 + 0^2 + (0-2)^2 = 4$. The distance is $\sqrt{4} = 2$.
• For points where $x^2+y^2 = \frac{3}{2}$ and $z=\frac{3}{2}$: $f(x,y,z) = (x^2+y^2) + (z-2)^2 = \frac{3}{2} + \left(\frac{3}{2}-2\right)^2 = \frac{3}{2} + \left(-\frac{1}{2}\right)^2 = \frac{3}{2} + \frac{1}{4} = \frac{6}{4} + \frac{1}{4} = \frac{7}{4}$. The distance is $\sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$.

Comparing the distances, $2 \approx 1.414$ and $\frac{\sqrt{7}}{2} \approx \frac{2.646}{2} \approx 1.323$.
The minimum distance is $\frac{\sqrt{7}}{2}$.

The points on the surface closest to $(0,0,2)$ are $\boxed{\text{all points }(x,y,z) \text{ such that } x^2+y^2 = \frac{3}{2} \text{ and } z=\frac{3}{2}}$.

We want to minimize/maximize the square of the distance from the origin:

The two constraints are:

We need to solve $\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2$.

The system of equations is:

1. $2x = \lambda_1(2x-4) + \lambda_2(-1)$
2. $2y = \lambda_1(2y) + \lambda_2(2y)$
3. $2z = \lambda_1(2z)$
4. $x^2 + y^2 + z^2 - 4x + 1 = 0$
5. $y^2 - x + 1 = 0$

From (3): $2z = 2\lambda_1 z \implies 2z(1-\lambda_1) = 0$. This implies $z=0$ or $\lambda_1=1$.

• Case 1: $z=0$ Substitute $z=0$ into (4): $x^2 + y^2 - 4x + 1 = 0$. From (5): $y^2 - x + 1 = 0 \implies y^2 = x - 1$. Substitute $y^2 = x-1$ into $x^2 + y^2 - 4x + 1 = 0$: $x^2 + (x-1) - 4x + 1 = 0$ $x^2 - 3x = 0$ $x(x-3) = 0$ This gives $x=0$ or $x=3$. If $x=0$, then $y^2 = 0-1 = -1$, which has no real solutions for $y$. If $x=3$, then $y^2 = 3-1 = 2 \implies y = \pm\sqrt{2}$. So, two candidate points are $(3, \sqrt{2}, 0)$ and $(3, -\sqrt{2}, 0)$.

• Case 2: $\lambda_1=1$ Substitute $\lambda_1=1$ into (2): $2y = 1(2y) + \lambda_2(2y) \implies 2y = 2y + 2\lambda_2 y \implies 2\lambda_2 y = 0$. This implies $\lambda_2=0$ or $y=0$. If $\lambda_2=0$: Substitute $\lambda_1=1, \lambda_2=0$ into (1): $2x = 1(2x-4) + 0(-1) \implies 2x = 2x-4 \implies 0 = -4$. This is a contradiction, so $\lambda_2 \ne 0$. Therefore, we must have $y=0$. Substitute $y=0$ into (5): $0^2 - x + 1 = 0 \implies x=1$. Substitute $x=1, y=0$ into (4): $1^2 + 0^2 + z^2 - 4(1) + 1 = 0 \implies 1 + z^2 - 4 + 1 = 0 \implies z^2 - 2 = 0 \implies z = \pm\sqrt{2}$. So, two more candidate points are $(1, 0, \sqrt{2})$ and $(1, 0, -\sqrt{2})$.

• For $(3, \sqrt{2}, 0)$: $f(3, \sqrt{2}, 0) = 3^2 + (\sqrt{2})^2 + 0^2 = 9 + 2 + 0 = 11$. The distance is $\sqrt{11}$.
• For $(3, -\sqrt{2}, 0)$: $f(3, -\sqrt{2}, 0) = 3^2 + (-\sqrt{2})^2 + 0^2 = 9 + 2 + 0 = 11$. The distance is $\sqrt{11}$.
• For $(1, 0, \sqrt{2})$: $f(1, 0, \sqrt{2}) = 1^2 + 0^2 + (\sqrt{2})^2 = 1 + 0 + 2 = 3$. The distance is $\sqrt{3}$.
• For $(1, 0, -\sqrt{2})$: $f(1, 0, -\sqrt{2}) = 1^2 + 0^2 + (-\sqrt{2})^2 = 1 + 0 + 2 = 3$. The distance is $\sqrt{3}$.

Comparing the distances: $\sqrt{11} \approx 3.317$ and $\sqrt{3} \approx 1.732$.
The maximum distance is $\sqrt{11}$.
The minimum distance is $\sqrt{3}$.

The maximum distance is $\boxed{\sqrt{11}}$ and the minimum distance is $\boxed{\sqrt{3}}$.