Let PQR be a triangle. Take a point A on or inside the triangle. Let f(x, y) = ax + by + c. Show that

**Discussion:**

Basic idea is this: First we take A on a side, say PQ. We show

Next we take A in the interior. Lets join RA and produce it to meet PQ at T. Using the previous argument we show

Hence it is sufficient to show for a point A on the line PQ (and the rest will follow):

Let Since A is on PQ, it is possible to write

Suppose f(A) is larger than both f(P) and f(Q). Then f(A) – f(P) and f(A) – f(Q) are both positive.

Let

Similarly we can show

Hence f(A) – f(Q) and f(A) – f(P) are opposite signs. The conclusion follows.

THIS CAN BE SOLVABLE USING PARTIAL DIFFERENTIATION

Can’t we do this by using the theorem of Linear Programming. When we subject the pt A to the required constraints, and maximise ax+by+c, by the theorem we get the maximum value at hte corner pts i.e P,Q,R

That might be possible. Can you send a detailed solution? We may publish it (with your name as author), if it is good.