Find the minimum value of $P = x^2 +y^2 - 4x + 8y \quad (x, y \in\mathbb {R}, x^2+ y^2 \leq 5)$

Question

The correct answer is $-15$.

This is how I ... (hopefully !) correctly solved it.

I rewrote the equation as ${(x-2)}^2 + (y+4)^2 = P + 20$

We also have ${x}^2 + {y}^2 \le 5$

I reasoned that $P = P_{min}$ when $(P + 20) = {(P + 20)}_{min}$

But $(P + 20) = r^2, \quad r \ge 0$

So $r^2 = r^2_{min}$ when $r = r_{min}$

I graphed the two circles and found that $r = r_{min}$ when the circles are tangent to each other, or when $r = \sqrt{5}$. From there, I found $P_{min} = -15$

My questions:

1. Are there any other ways to find $P_{min}$? (e.g. without using graphs)
2. I first did it this way and got $-11$, which is incorrect. But I'm not sure what goes wrong.

$P = x^2 - 4x + y^2 + 8y$

$P = P_{min}$ when $x = \frac{-b}{2a} = 2$

$P_{min} = y^2 + 8y - 4$

$x^2 + y^2 \le 5$

$x = 2 \implies y^2 \le 1 \iff -1 < y < 1$

Graph $P_{min}$, we see that on $[-1;1]$, $P_{min}$ is smallest when $P_{min} = -11$ and $y = -1$

Thank you!

Answer 1

\begin{align} P & = x^2+y^2-4x+8y \\ & = (x-2)^2+(y+4)^2-20 \\ & = \left( \sqrt{(x-2)^2+(y+4)^2} \right)^2 - 20 \end{align}

Thus minimizing $P$ subject to $x^2+y^2 \leq 5 $ is equivalent to minimizing the distance of $A(2,-4)$ from any point on or inside the circle $x^2+y^2=5$.

As is seen from the figure, the minimum distance is attained at $B(1,-2)$.

Hence minimum $P = (1-2)^2+(-2+4)^2-20 = -15. $

Answer 2

One possible solution is to parametrize $(x,y)$ as follows: $$ x = s\cos \varphi \\ y = s\sin \varphi $$ with $s\in \left[0,\sqrt{5}\right]$. Then $$ P = s^2 -4s\cos\varphi +8s\sin\varphi = s^2 + 4\sqrt{5}s\cos\left(\varphi-\varphi_0\right) $$ with $\varphi_0$ defined by the equations $\cos\varphi_0 = -\frac{1}{\sqrt{5}}$ and $\sin\varphi_0 = \frac{2}{\sqrt{5}}$

It is obvious that the minimum will be attained when $\cos\left(\varphi-\varphi_0\right) = -1.$ Therefore, we only have to check which valid value of $s$ minimizes $s^2-4\sqrt{5}s.$ This function is decreasing in the interval $\left[0,\sqrt{5}\right]$. Therefore, the solution is $s=\sqrt{5}.$ This leads to $P=-15.$

Answer 3

First, we should check if the global minimum of $ x^2 + y^2 - 4x + 8 y $ is inside the disc $x^2 + y^2 \le 5 $.

By completing the square in $x$ and $y$, we get

$x^2 + y^2 - 4 x + 8 y = (x - 2)^2 - 4 + (y + 4)^2 - 16 = (x-2)^2 + (y+4)^2 - 20$

The global minimum is therefore at $(2, -4)$, but $2^2 + (-4)^2 \gt 5 $, so the minimum lies on the circumference of $x^2 + y^2 = 5$

Now define the Lagrange multiplier function

$ g(x, y) = x^2 + y^2 - 4 x + 8 y + \lambda ( x^2 + y^2 - 5 ) $

Then

$ g_x = 2 x - 4 + \lambda (2 x ) = 0 $

$ g_y = 2 y + 8 + \lambda (2 y ) = 0 $

$ g_\lambda = x^2 + y^2 - 5 = 0 $

From the first two equations, we have

$ \lambda = \dfrac{ 2 - x }{x} = \dfrac{- y - 4 }{y} $

Cross multiply,

$ (2 - x) y = x (- y - 4) $

From which

$ 2 y = - 4 x $ or $ y = -2 x $

Substitute this into $g_\lambda = 0 $

$ x^2 + 4 x^2 = 5 $

So $ x = \pm 1 $ and the corresponding $y$ is $ y = \mp 2 $

So we have two possible solutions $(1, -2)$ or $(-1, 2)$.

Now, we just need to compute the value of the function at each point and compare the values.

$ f(1, -2) = (1)^2 + (-2)^2 - 4(1) + 8(-2) = -15 $

$ f(-1, 2) = (-1)^2 + (2)^2 - 4(-1) + 8(2) = 25 $

So the minimum is $-15$ at $(1, -2)$, and the maximum is $25$ at $(-1, 2) $.

Answer 4

Using CS inequality, $ 5\cdot 80\geqslant (x^2+y^2)(4^2+8^2)\geqslant (-4x+8y)^2$. Let $z=-4x+8y$, so $z^2 \leqslant 20^2 \implies -20\leqslant z \leqslant 20$.

Therefore consider $80P = (x^2+y^2)(4^2+8^2)+80(-4x+8y) \geqslant z^2+80z = (z+40)^2-40^2 \\ \geqslant (-20+40)^2-40^2=-1200 \implies P\geqslant -15$

All the inequalities above become equalities simultaneously when $x^2+y^2=5$, $z=-20=-4x+8y$ and $x:y=1:-2$, i.e. when $(x, y)=(1, -2)$, hence this is the minimum.

Answer 5

Recall that, you can also reach the solution immediately by completing the square method .

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$$ \begin{align}P(x,y):&=x^2+y^2-4x+8y\\ &=2(x-1)^2+2(y+2)^2\\ &\,\,\,\,\;+\underbrace{\color{#c00}{\left(\color{black}{\left(5-(x^2+y^2\right)}\right)}}_{\color{#c00}{≥\thinspace 0}}-15\\ &≥-15\;\;\;\tiny{\blacksquare}\end{align} $$

Equality holds iff, when $x-1=0$ and $y+2=0$ and $x^2+y^2=5$, which is equivalent to $(x,y)=(1,-2)\thinspace.$