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About this question I have asked lately $x^2+y^2 \leq 16$ and also $\{x,y\} \subset \mathbb{R}^+$ what is the minimum value of $3x+4y$.

I want to solve it with Lagrange now, but I couldn't manage to. I have tried to construct a multi-variable function which was the following $f(x,y)=3x+4y$ then to find the minimum value of it, but I am really not accustomed to using derivatives for extremum problems. And I couldn't find the connection with it between the Lagrange multiplier. What do you suggest?

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Let $K=\{(x,y) \in \mathbb R^2:x^2+y^2 \le 16\}$. We observe that there no points$(x_0,y_0)$ such that $f_x(x_0,y_0)=0$ and $f_y(x_0,y_0)=0$.

Consequence: $\min f(K)= \min f( \partial K)$

Hence you have to minimize the function $f$ under the condition $x^2+y^2=16$.

Can you proceed from here ?

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  • $\begingroup$ So what I have to do is $f(x,y,k)=3x+4y+k(x^2+y^2-16)$ ? $\endgroup$ – Deniz Tuna Yalçın Sep 7 '17 at 9:17
  • $\begingroup$ I couldn't find a decent solution using this: $3+2kx=4+2ky$ and I can't keep going $\endgroup$ – Deniz Tuna Yalçın Sep 7 '17 at 9:18
  • $\begingroup$ Yes, but since $f(x,y)=3x+4y$ you shoul write $g(x,y,k)=3x+4y+k(x^2+y^2-16)$. $\endgroup$ – Fred Sep 7 '17 at 9:19
  • $\begingroup$ I forgot that, but that didn't change the deriving process $g_x'=3+2kx$ and $g_y'=4+2ky$ and $g_k'=0$ so $3+2kx=4+2ky$. How should I proceed, or am I making a mistake? $\endgroup$ – Deniz Tuna Yalçın Sep 7 '17 at 9:21
  • $\begingroup$ You have to solve the system $3+2kx=0, \quad 4+2ky=0$ $\endgroup$ – Fred Sep 7 '17 at 9:22
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By C-S $$3x+4y\geq-|3x+4y|\geq-\sqrt{(3^2+4^2)(x^2+y^2)}\geq-\sqrt{25\cdot16}=-20.$$ The equality occurs for $(3,4)||(x,y)$ and $x^2+y^2=16$, which says that $-20$ is a minimal value.

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  • $\begingroup$ In the question text I have written $x,y \in \mathbb{R}^+$ $\endgroup$ – Deniz Tuna Yalçın Sep 7 '17 at 10:05
  • $\begingroup$ Although this will be helpful to evaluate negative expressions with C-S. Thank you:) $\endgroup$ – Deniz Tuna Yalçın Sep 7 '17 at 10:06
  • $\begingroup$ @Deniz Tuna Yalçın You are welcome! $\endgroup$ – Michael Rozenberg Sep 7 '17 at 10:07
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With Cauchy-Schwarz:

$|3x+4y| \le |(3,4)*(x,y)| \le ||(3,4)||*||(x,y)|| = 5*4=20$

hence:

$-20 \le 3x+4y$.

Now look for $(x,y)$ such that $-20 = 3x+4y$ and $x^2+y^2=16$.

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  • $\begingroup$ I recall writing $x,y \in \mathbb{R}^+$ although I have always known C-S to be a method only for positive numbers. Seeing this usage is helpful, thank you:) $\endgroup$ – Deniz Tuna Yalçın Sep 7 '17 at 10:08
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1) $x^2 + y^2 = 16,$ and interior $(\lt )$,

equation of a circle about $O(0,0)$ , and radius $= 4$.

2) Straight line: $3x + 4y = C $, or

$y = -(3/4)x + C/4$.

Slope $m= -3/4$; $Y -$ intercept: $C/4.$

We are looking for a straight line with minimal $Y-$intercept $C/4$.

Three cases:

A) The line does not intersect the circle. Ruled out.

2) The line intersects the circle twice. We can still move in neg. $Y-$ direction to decrease $C/4$

3) The line touches the circle (tangent line).

There are 2 Points where the line is tangential to the circle:

$P_1$, $P_2 $, and

distance $ OP_1 = OP_2 = 4$.

Using the distance formula for $3x + 4y - C = 0$:

$4 = \frac{-C}{^+_- \sqrt{3^2+4^2}}$

Hence:

$C_{min} = -20,$ $C_{max} = 20$.

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