Help with minimization problem

Question

help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?

Answer 1

Then it's a parabola, which I suspect you know how to minimize. (Or you can use calculus if you want to).

Answer 2

The quantity $x^2+y^2$ is minimized when $\sqrt{x^2+y^2}$ is minimized, and $\sqrt{x^2+y^2}$ is just the distance from the point $\langle x,y\rangle$ to the origin. Thus, you’re looking for the point on the line $3x-4y=12$ that is closest to the origin. Call this line $\ell$; the point closest to the origin is the point of intersection of $\ell$ with a line through the origin and perpendicular to $\ell$. The slope of $\ell$ is $\frac34$, so the slope of the perpendicular is $-\frac43$: you want the intersection of $\ell$ and the line $y=-\frac43x$.

Answer 3

We start from your expression for $z$ in terms of $y$. The standard "calculus" way of handling the problem is to calculate $\frac{dx}{dy}$, and set it equal to $0$ to find the critical points.

For the calculation of $\frac{dz}{dy}$, you can first simplify, and then differentiate, or differentiate, and then simplify.

I will do it one way, and you can do it the other way. We have $$z=\frac{16}{9}(y+3)^2+y^2.$$ Differentiate. We get $$\frac{dz}{dy}=\frac{32}{9}(y+3)+2y=\frac{1}{9}(50y+96).$$ Set this equal to $0$ and solve for $y$.

Don't forget to check that we really do get a minimum.

Answer 4

To answer your question, "and then?", with details, this is a way the calculation you began could continue, and this is without calculus:

$z=(4\frac{y+3}{3})^2 + y^2 = 16\frac{(y+3)^2}{9} + \frac{9y^2}{9} = \frac{16(y^2+6y+9)+9y^2}{9} = \frac{25}{9}y^2 + \frac{32}{3}y + 16$.

Center of a parabola is at $y = \frac{-b}{2a} = \frac{-32/3}{50/9} = -\frac{48}{25}$.

Finally, $x=4\frac{y+3}{3} = \frac{4(27/25)}{3} =\frac{36}{25}$.

The method described by Brian Scott is a bit easier, I think.

Answer 5

$$ {\rm F} \equiv x^{2} + y^{2} -\mu\left(3x - 4y - 12\right)\,, \quad \left\vert \begin{array}{rcl} {\partial F \over \partial x} = 0 &\Longrightarrow& 2x - 3\mu = 0\ \Longrightarrow\ x = {3 \over 2}\,\mu \\[1mm] {\partial F \over \partial y} = 0 &\Longrightarrow& 2y + 4\mu = 0\ \Longrightarrow\ y = -2\mu \end{array}\right. $$

$$ 12 = 3x - 4y = {9 \over 2}\,\mu + 8\mu = {25 \over 2}\,\mu\ \Longrightarrow\ \mu = {24 \over 25} \quad\Longrightarrow\quad \left\vert% \begin{array}{rcl} x & = &{3 \over 2}\,{24 \over 25} = {36 \over 25} \\[2mm] y & = & -2\,{24 \over 25} = -\,{48 \over 25} \end{array}\right. $$

$$ \begin{array}{|c|}\hline\\ {\large\quad% z_{\rm min} = \left(36 \over 25\right)^{2} + \left(-\,{48 \over 25}\right)^{2} = \color{#0000ff}{144}\quad} \\ \\ \hline \end{array} $$