I'm asked to find the minimum and maximum values of $f(x, y, z) = x^2+y^2+z^2$ given the constraints $x+2y+z=3$ and $x-y=7$.
I'm pretty sure I need to set up Lagrange equations, giving:
$$2x = A+B$$ $$2y = 2A-B$$ $$2z = A$$
From here, I'm not sure where to go.
Well Lagrange multiplier will help you, but since you have 2 equations, you can easily to reduce the function to a one variable, which is easily to maximize or minimize. So from the two equations, you have:
$$x=y+7; \quad \text{and} \quad$$ $$x+2y + z = 3 \iff y+7+2y+z=3 \iff z=-4-3y$$
Now you have:
$$f(x,y,z) = x^2 + y^2 + z^2 \iff f(x)=(y+7)^2 + y^2 + (-4-3y)^2$$
$$f(x) = y^2 + 14y + 49 + y^2 + 16 + 24y + 9y^2 = 11y^2 + 38y + 65$$
Now find the derviative, set it to 0 and find the extreme point:
$$f'(x) = 22y + 38 = 0 \implies y= -\frac{19}{11}$$
Obviously this is the minimum of the function. So the minimum occurs at:
$$f\left(\frac{58}{11}, -\frac{19}{11}, -\frac{13}{11}\right)= \frac{3894}{121} = 32.(18)$$
On the other side this function doesn't have maximum. Let $y \to \infty$, from this we have: $x \to \infty$ and $z\to -\infty$, obviouslt $x^2 + y^2 + z^2 \to \infty$