Let PQR be a triangle. Take a point A on or inside the triangle. Let f(x, y) = ax + by + c. Show that \(\mathbf { f(A) \le \max { f(P), f(Q) , f(R)} }\)

**Discussion:**

Basic idea is this: First we take A on a side, say PQ. We show \(\mathbf { f(A) \le \max { f(P), f(Q) } \implies f(A) \le \max { f(P), f(Q) , f(R)} }\)

Next we take A in the interior. Lets join RA and produce it to meet PQ at T. Using the previous argument we show \(\mathbf { f(A) \le \max { f(T), f(R) } \text{but} f(T) \le max { f(P), f(Q) } \implies f(A) \le \max { f(P), f(Q) , f(R)} }\)

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

Let \(\mathbf { A = (x, y), P = (x_p , y_p) , Q = (x_q, y_q) }\) Since A is on PQ, it is possible to write \(\mathbf { x = \frac{m x_p + n x_q }{ m+n} , y = \frac{ m y_p + n y_q} {m+n}, m,n in \mathbb{R}^{+} }\)

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 \(\mathbf { \lambda_1 = \frac{m}{m+n} , \lambda_2 = \frac{n}{m+n} implies \lambda_1 + \lambda_2 = 1 }\)

\(\mathbf { f(A) – f(P) = ax+by+c – ax_p – by_p -c = a(x-x_p) + b(y – y_p) = \newline a(\lambda_1 x_p + \lambda_2 x_q – x_p) + b (\lambda_1 y_p + \lambda_2 y_q – y_p) }\)

\(\mathbf {\implies f(A) – f(P) = a(\lambda_2 x_q – (1-\lambda_1) x_p) + b(\lambda_2 y_q – (1-\lambda_1) y_p) = a \lambda_2 (x_q – x_p) + b \lambda_ 2 (y_q – y_p) }\)

Similarly we can show \(\mathbf { f(A) – f(Q) = a \lambda_1 (x_p – x_q) + b \lambda_1 (y_p – y_q) }\)

\(\mathbf { f(A) – f(Q) = \frac{-\lambda_2}{\lambda_1} (f(A) – f(P) ) }\)

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.