Let $L ( x , y ) = a x + b y$ .
There exist values of a and b so that $L ( x , y )$ takes a minimum value of 8 on the unit circle $x ^ { 2 } + y ^ { 2 } = 1$ at the point $\left( \frac { 1 } { 2 } , \frac { - \sqrt { 3 } } { 2 } \right)$ .

Question

Let $L ( x , y ) = a x + b y$ .
 There exist values of a and b so that $L ( x , y )$ takes a minimum value of 8 on the unit circle $x ^ { 2 } + y ^ { 2 } = 1$ at the point $\left( \frac { 1 } { 2 } , \frac { - \sqrt { 3 } } { 2 } \right)$ .

Quizplus · Accepted Answer

The function $L(x, y) = ax + by$ is linear, and on the unit circle, it cannot have a minimum value at a specific point unless it is constant, which is not possible for a linear function on a circle.

Let L(x,y)=ax+byL ( x , y ) = a x + b yL(x,y)=ax+by

Let $L ( x , y ) = a x + b y$