# How do I find the minimum and maximum of a multivariable function given two constraints?

Find the minimum and maximum of $$f(x, y, z) = y + 4z$$ subject to two constraints, $$3x + z = 5$$ and $$x^2 + y^2 = 1$$.

Having a hard time figuring out how to do this problem. I think I'm doing it right but I can't seem to get the correct answer in exact terms.

Here's what I got so far (the labels are respectively up and down is left to right so $$g(x,y,z)=3x+z=5$$:

$$f_x = g_xλ + h_x\delta$$ $$f_y = g_yλ + h_y\delta$$ $$f_z = g_zλ + h_z\delta$$

$$0 = (3)λ + (2x)\delta$$ $$1 = (0)λ + (2y)\delta$$ $$4 = (1)λ + (0)\delta$$

After doing the calculations I get $$λ = 4$$.

Now I solve for $$x$$ and $$y$$ by plugging in that value into the equations.

I get:

$$x=-\frac{6}{\delta}$$ $$y=\frac{1}{2\delta}$$

Then I plug it into $$f(x,y,z)$$ and get:

$$\frac{6}{\delta}^2+\frac{1}{2\delta}^2=1$$

$$\delta = \frac{\sqrt{(145)}}{2}$$

So solving for $$x$$ and $$y$$ again:

$$x=-\frac{-12}{\sqrt{145}}$$ $$y=\frac{1}{\sqrt{145}}$$

Then I solve for $$z$$ by using $$g(x,y,z)$$:

$$-\frac{-12}{\sqrt{145}}*3+z=5$$ $$z=5+\frac{36\sqrt{145}}{145}$$

So then my point is: $$(\frac{12}{\sqrt{145}},\frac{1}{\sqrt{145}},5+\frac{36\sqrt{145}}{145})$$

I plug this into $$f(x,y,z)$$:

$$f(x,y,z)=\frac{1}{\sqrt{145}}+4\left(5-\frac{36\sqrt{145}}{145}\right)$$

and get: $$f(x,y,z)=\sqrt{145}+20$$

And so I use the opposite point to get the other value (each value multiplied by -1):

$$f(x,y,z)=-\frac{1}{\sqrt{145}}+4\left(5-\frac{36\sqrt{145}}{145}\right)$$

$$f(x,y,z)=-\sqrt{145}+20\quad$$

$$maximum = \sqrt{145}+20\quad$$ $$minimum = -\sqrt{145}+20\quad$$

Yet they're both wrong. I have no clue what's happening. I've checked my calculations a lot of times. I must be missing steps somewhere.

I don't know how to solve this problem.

If you just want to provide a final answer that'll at least help me back track. Thank you.

• First, I would recommend avoiding solving for $\delta$. Instead, combine your equations to eliminate $\delta$ and yet $x=-12y$. Second, and more important, you lost a root when you solve the equations. In particular, $y=\pm 1/\sqrt{145}$. I think you messed up a sign, since you have $x=+12y$, not $x=-12y$. Regardless, the arithmetic is yuck. – Ted Shifrin Oct 10 '20 at 22:37
• @TedShifrin thanks for the feedback. – Si Random Oct 10 '20 at 22:39
• I see. You didn't lose a root. But your exposition is mathematically sloppy. At the end, you just took the negative of the solution you had. Be careful; this doesn't always work. – Ted Shifrin Oct 10 '20 at 22:47
• @TedShifrin i retried with your suggestion and i'm still getting the wrong answer. I don't know what's happening. I spent 5 hours on this one problem and i'm so desperate. – Si Random Oct 10 '20 at 22:49
• $y=\pm 1/\sqrt{145}$, $x=-12y$, $z=5-3(-12y)=5+36y$. So we get $\big(-\frac{12}{\sqrt{145}},\frac1{\sqrt{145}},5+\frac{36}{\sqrt{145}}\big)$. Aha, note that as I cautioned you, you can't just take the negative. The other solution is $\big(\frac{12}{\sqrt{145}},-\frac1{\sqrt{145}},5-\frac{36}{\sqrt{145}}\big)$, and the $z$ coordinate is far from the negative. The $f$-values are $y+4z=y+4(5+36y)=20+145y$, very conveniently. So we get $20\pm\sqrt{145}$, nice and simple. I see that you've now gotten this as well. So what's the problem? ... Note that the final formula makes things pretty easy. – Ted Shifrin Oct 10 '20 at 23:40

Since the objective function is a function of $$y,z$$ can we rewrite the constraints to eliminate x, and have objective and constraints in the same variables?

$$3x + z = 5\\ x = \frac {5-z}{3}\\ x^2 + y^2 = 1\\ \left(\frac {5-z}{3}\right)^2 + y^2 = 1$$

We have an ellipse. We need to find where the tangent of the ellipse is parallel to $$y+4z$$

$$-2\frac {5-z}{9}\ dz + 2y\ dy = 0\\ \frac {dy}{dz} = \frac {5-z}{9y}\\ \frac {dy}{dz} = -4\\ 5-z = -36y$$

and plug this back into our constraint.

$$145 y^2 = 1\\ y = \pm \frac {1}{\sqrt {145}}\\ z = 5 \pm \frac {36}{\sqrt{145}}\\ f(x,\frac {1}{\sqrt {145}},5+\frac {36}{\sqrt{145}}) = 20 + \sqrt{145}\\ f(x,-\frac {1}{\sqrt {145}},5-\frac {36}{\sqrt{145}}) = 20 -\sqrt{145}$$

Which is the same as you have above.

Otherwise, we could do something with Lagrange multipliers

$$F(x,y,z,\lambda,\mu) = y+4z + \lambda (x^2 + y^2 - 1) + \mu (3x + z -5)\\ \frac {\partial F}{\partial x} = 2\lambda x + 3\mu = 0\\ \frac {\partial F}{\partial y} = 1 + 2\lambda y = 0\\ \frac {\partial F}{\partial z} = 4 + \mu = 0\\ \frac {\partial F}{\partial \lambda} = x^2+y^2 - 1 = 0\\ \frac {\partial F}{\partial \mu} = 3x+z - 5 = 0$$

And solve.

• my answer was wrong because I had them in the wrong slots... also thank you so much. I am forever in your debt. this is much faster approach. i tried 5 discord servers and so many other places and no one could help me so thank you so much – Si Random Oct 10 '20 at 23:07