A diet needs at least 10 units of fibre and at least 15 units of iron…

Step 1. The costs in terms of $p$. At $(0, 15)$, $Z = 60$. At $(4, 3)$, $Z = 4p + 12$. At $(10, 0)$, $Z = 10p$.

Step 2. The first tie. Setting $4p + 12 = 10p$ gives $6p = 12$, so $p = 2$. Then each diet costs $20$ rupees. For $p$ below $2$, the all-P diet at $(10, 0)$ is cheaper.

Step 3. The second tie. Setting $4p + 12 = 60$ gives $4p = 48$, so $p = 12$. For $p$ above $12$ the diet $(0, 15)$ is cheaper.

Step 4. Say it. The mixed diet of $4$ kg of P and $3$ kg of Q is cheapest when $2 \le p \le 12$.