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.
$\endgroup$2 Answers
$\begingroup$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 $x^2 + y^2 + z^2 \to \infty$
$\endgroup$ $\begingroup$Here's how to do it with Lagrange multipliers.
Put $g(x,y,z) = x + 2y + z -3 = 0$ and $h(x,y) = x - y - 7 =0$ as constraints.
Then we have 5 equations
- $f_x = 2x = \lambda g_x + \mu h_x = \lambda + \mu$
- $f_y = 2y = \lambda g_y + \mu h_y = 2\lambda - \mu$
- $f_z = 2z = \lambda g_z + \mu h_z = \lambda + 0 = \lambda$
- $g(x,y,z) = x + 2y + z -3 = 0$
- $h(x,y) = x - y - 7 =0$
Now, from (3) we have $\frac{\lambda}{2} = z$. Using our expressions for $x$ and $y$ from (1) and (2) and substituting them in in (4) and (5),
we have now
- $\frac{\lambda + \mu}{2} + 2\lambda - \mu + \frac{\lambda}{2} -3 = 3\lambda - \frac{\mu}{2} -3 = 0$
- $\frac{\lambda + \mu}{2} - \lambda + \frac{\mu}{2} - 7 = -\frac{\lambda}{2} + \mu -7 = 0$.
Now we have two linear equations, two unknowns. Multiply the first by two and add to the second two get
- $\frac{11 \lambda}{2} -13 =0 \rightarrow \lambda = \frac{26}{11}$.
Plugging this into the first equation above, $\mu = \frac{90}{11}$.
Now from (1), (2), (3), we see that our only critical point occurs at the point where $x = \frac{58}{11}, y = \frac{-19}{11}, z= \frac{13}{11}$.
Plugging this into our equation for $f$, this comes out to $f(\frac{58}{11}, \frac{-19}{11}, \frac{13}{11}) = \frac{3894}{121} = \frac{354}{11}$.
Now, how do we know whether this is the minimum or maximum? To figure it out we just have to produce a point whose function value is less than or greater than $\frac{354}{11}$. If there's one with a higher function value, than this must be a minimum, and if there's one with a lower function value it's a maximum.
We want points satisfying (4) and (5). For ease, let $z=0$. Then we just pick the point of intersection of the lines $ y = \frac{-x}{2} + \frac{3}{2}$ and $y = x -7$. This point is $(\frac{17}{3}, \frac{-4}{3})$.
Plugging this into $f$, we see $f(\frac{17}{3}, \frac{-4}{3}, 0)= \frac{305}{9} \gt \frac{354}{11} = f(\frac{58}{11}, \frac{-19}{11}, \frac{13}{11})$. So our point must be a minimum.
Then we must have the minimum of $f$ is $f(\frac{58}{11}, \frac{-19}{11}, \frac{13}{11}) = \frac{3894}{121} = \frac{354}{11}$.
$\endgroup$
