Calculus 3: Lagrange Multipliers

Question

Find the minimum and maximum of $f(x,y,z) = x^2+y^2+z^2$ subject to two constraints, $x+2y+z=8$ and $x−y=5$.

Looking at the equation, it's clear that there is no maximum.

After working this problem out, I found: $x = 41/11$ , $y = -14/11$ , and $z = 9/11$

After plugging this into the original equation, I found the minimum to be $178/11$

However, my online homework is saying my answer is incorrect. Did I do something wrong?

Thank you in advance to anyone who can help me out with this.

Answer 1

To check the result use that from the constraints

• $y=x-5$

• $z=8-x-2y=18-3x$

then we need to find the extrema for

$$g(x)=f(x,x-5,18-3x)=x^2+(x-5)^2+(18-3x)^2=11x^2-118x+349$$

Answer 2

I solved equation, and got an answer of $x=\dfrac{23}{3}$ ,$y=\dfrac{8}{3}$,$z=-5$. Is it the correct answer, by your online homework? If it is, I will share my solution with you. By the way, your answer doesn't fit the first constraint, you can check it by plugging in values.