About using lagrange multiplier for the min value of $3x+4y$ when $x^2+y^2\leq 16$ and $x,y \in \mathbb{R}$

Question

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?

Answer 1

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 ?

Answer 2

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.

Answer 3

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 $x^2+y^2=16$.

Answer 4

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:

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