A hospital kitchen mixes two foods, P and Q…

Step 1. The model. Minimise $Z = 5x + 4y$ rupees subject to $x + 2y \ge 10$, $3x + y \ge 15$, $x \ge 0$ and $y \ge 0$.

Step 2. The costs. At $(0, 15)$, $Z = 60$. At $(4, 3)$, $Z = 20 + 12 = 32$. At $(10, 0)$, $Z = 50$. The smallest is $32$.

Step 3. The test. The line $5x + 4y = 32$ has slope $-\dfrac{5}{4}$. The edges at $(4, 3)$ have slopes $-3$ and $-\dfrac{1}{2}$. Since $-3 < -\dfrac{5}{4} < -\dfrac{1}{2}$, the line touches the region only at $(4, 3)$. So $32$ is the minimum.

Step 4. Say it. Use $4$ kg of P and $3$ kg of Q. The cost is $32$ rupees.