Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plane $x+y+z=1$

Question

Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plain $x+y+z=1$

Hi i was doing this question but i'm not sure i was right. Does this make sense?

Let

$$f(x) = x^2+y^2 = 1$$ $$f(y) = x+y+z=1, \quad \text{our constraint}$$

$$ \nabla f(x) = \lambda \nabla g(x) $$

$f_x = 2x = \lambda$ ..... (1)

$f_y = 2y = \lambda$ ...... (2)

$f_z = 0 = \lambda$

(1) = (2)

$2x = 2y = \lambda = 0 $

$x=y$

Sub into f(x)

$$f(x) = x^2 + x^2 = 1$$ $$f(x) = 2x^2 = 1$$ $$f(x) = x = \pm \frac{1}{\sqrt{2}} $$ $$ \therefore \, y = \pm \frac{1}{\sqrt{2}}$$

Points: $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right), \quad \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)$

Sub these points into g

$$g\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, z \right) = \frac{2}{\sqrt{2}} + z = 10 $$ $$ z = 1 - \frac{2}{\sqrt{2}} $$

$$g\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, z \right) = -\frac{2}{\sqrt{2}} + z = 10 $$ $$ z = 1 + \frac{2}{\sqrt{2}} $$

So highest and lowest points are.... $$\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1 - \frac{2}{\sqrt{2}} \right) \quad lowest$$ $$\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 1 + \frac{2}{\sqrt{2}} \right) \quad highest$$

Answer 1

By inequality $2x^2+2y^2\geq (x+y)^2$ we indeed have $-\sqrt2 \leq x+y \leq \sqrt{2}$ and to maximize/minimize $z$ we need to minimize/maximize $x+y$ so your solution until $x=y=\pm{1\over\sqrt2}$ is correct.

However how do you get $z$ looks unreasonable as $x+y+z=1$ but your solution, for example ${1\over\sqrt2}+{1\over\sqrt2}+{1\over\sqrt8}$ does not equal $1$.

You should simply calculate $z=1-x-y=1+{\sqrt2}$ for maximum and $1-\sqrt2$ for minimum.

There are two major mistakes after a look at your way of finding z:

(1) $g(x,y,z)=1\neq10$ as stated in the question. I do not know where the $10$ comes from.

(2) $10-{2\over \sqrt2}$ is not $8\over\sqrt2$. You cannot calculate irrational numbers like this.

As to answer the comment, what Matthew is talking about is the general way of solving multiple constraints.

We want to maximize $h(x,y,z)=z$ using constraint $g(x,y,z)=x+y+z=1$ and $f(x,y,z)=s^2+y^2=1$. Let the first multiplier be $\lambda$ and the second be $u$

Then we have the system:

$0=\lambda+2ux=\lambda+2uy$

$1=\lambda$

$x^2+y^2=1$

$x+y+z=1$

By solving the equations we will find the value of $z$.

Answer 2

No, this is not correct. It looks like you assumed that the first function described in the problem was the objective function (that which is to be optimized) and the second function is the constraint function.

But if you read the problem, you see that both $x^2 + y^2 = 1$ and $x+y+z=1$ are constraints. So you need a Lagrange multiplier approach with two multipliers.

But wait! If these are both constraints, what is the objective function? Read it again. What words in the problem say “this is an optimization problem.” Optimal means superlative, and the superlative words are highest and lowest.

What function would have its maximum value achieved at the highest point of a set? The height function $f(x,y,z) = z$.

---

Here is a fleshed-out solution. The Lagrange equation is $$ \nabla (z) = \lambda \nabla (x^2 + y^2) + \mu \nabla (x+y+z) $$ In coordinates, \begin{align} 0 &= \lambda(2x) + \mu \tag{1}\\ 0 &= \lambda(2y) + \mu \tag{2}\\ 1 &= \mu \tag{3} \end{align} Then we have the equations of constraint \begin{align} x^2 + y^2 = 1 \tag{4} \\ x + y + z = 1 \tag{5} \end{align}

Substituting (3) into (1) and (2) gives \begin{align*} x =y= -\frac{1}{2\lambda} \end{align*} Then by (4), $$ \left(-\frac{1}{2\lambda}\right)^2 + \left(-\frac{1}{2\lambda}\right)^2 = 1 \implies \lambda^2 = \frac{1}{2} \implies \lambda = \pm \frac{1}{\sqrt{2}} $$ If $\lambda = \frac{1}{\sqrt{2}}$, then $x = y = -\frac{1}{\sqrt{2}}$ and by (5) $z=1+\sqrt{2}$. If $\lambda = - \frac{1}{2}$ then $x=y=\frac{1}{\sqrt{2}}$ and $z=1-\sqrt{2}$. So the highest point is $\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},1+\sqrt{2}\right)$ and the lowest point is $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},1-\sqrt{2}\right)$