Diet and mixing problems

What you will be able to do

• write the constraints and the cost of a diet problem;
• see why its region is unbounded and why a minimum still exists;
• find the cheapest mix and say how its price changes when a food gets dearer or cheaper.

The shape of a diet problem

A canteen mixes two foods. Food P has $3$ units of protein and $1$ unit of iron in each kilogram. Food Q has $1$ unit of protein and $2$ units of iron in each kilogram. A meal needs at least $18$ units of protein and at least $16$ units of iron. Food P costs $6$ rupees a kilogram and food Q costs $5$ rupees a kilogram. What is the cheapest mix?

1. The unknowns. Let $x$ kg of P and $y$ kg of Q go into a meal.
2. The objective. The cost is $Z = 6x + 5y$ rupees. We want it as small as possible.
3. The constraints. Protein: $3x + y \ge 18$. Iron: $x + 2y \ge 16$. Also $x \ge 0$ and $y \ge 0$.

Notice the signs. The needs are minimums, so the signs are $\ge$. A resource limit in a factory was $\le$. In a diet it is the other way round. Eating more of a food never hurts the needs, so the region has no upper edge. It is unbounded.

Solve it

4. The corners. Multiply $3x + y = 18$ by $2$ to get $6x + 2y = 36$. Subtract $x + 2y = 16$ to get $5x = 20$, so the lines meet at $x = 4$, $y = 6$. The line $3x + y = 18$ meets the $y$ axis at $(0, 18)$. The line $x + 2y = 16$ meets the $x$ axis at $(16, 0)$. The corners are $(0, 18)$, $(4, 6)$ and $(16, 0)$.

5. The costs. At $(0, 18)$, $Z = 90$. At $(4, 6)$, $Z = 24 + 30 = 54$. At $(16, 0)$, $Z = 96$. The smallest is $54$.
6. The test. The region is unbounded, so check the half-plane $6x + 5y < 54$. The line $6x + 5y = 54$ has slope $-\dfrac{6}{5}$. The two edges at $(4, 6)$ have slopes $-3$ and $-\dfrac{1}{2}$. Since $-3 < -\dfrac{6}{5} < -\dfrac{1}{2}$, the line touches the region only at $(4, 6)$, and the half-plane has no feasible point.
7. The answer. The cheapest meal uses $4$ kg of P and $6$ kg of Q and costs $54$ rupees.

Check the needs. Protein is $3(4) + 6 = 18$ units and iron is $4 + 2(6) = 16$ units. Both are exactly met.

What if the price changes?

Suppose food P gets cheaper. At $2.5$ rupees a kilogram, the corner $(4, 6)$ costs $4(2.5) + 30 = 40$ rupees and the corner $(16, 0)$ costs $16(2.5) = 40$ rupees. They tie. Below $2.5$ rupees, the all-P diet at $(16, 0)$ is cheaper.

So the best diet changes when the prices change, and the corner where it changes is found by setting two corner costs equal.

Practise with new numbers

Try it in Python

The program works out the cost at each corner and asks linprog. The needs use $\ge$, so each row is multiplied by $-1$. Then it sweeps the price of food P.

Common mistakes

• Using $\le$ for a need. A need is a minimum, so it is $\ge$.
• Looking for a maximum cost. A diet problem asks for the minimum. The maximum does not exist on an unbounded region.
• Taking the smallest corner without the test. On an unbounded region, test that no feasible point is cheaper.
• Mixing up units. Say kilograms of each food and rupees for the cost.

Remember these

• Diet problems have $\ge$ constraints, a cost to minimise, and an unbounded region.
• Work out the cost at every corner. The smallest corner cost is the minimum when the prices are positive.
• If a price changes, the cheapest corner changes where two corner costs tie.
• Answer in the units of the story.

References

• NCERT, Mathematics Textbook for Class XII, Part II, Chapter 12, Linear Programming (diet problems).

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Food A has $2$ units of calcium in each kilogram and food B has $1$ unit. A meal needs at least $12$ units. Write the rule.

2. The corners of a diet region are $(0, 10)$, $(3, 4)$ and $(8, 0)$. Food A costs $4$ rupees a kilogram and food B costs $3$ rupees. Find the lowest cost.

3. Why does a diet problem have no maximum cost?

4. A diet uses $x$ kg of P and $y$ kg of Q. Food P costs $6$ rupees a kg and food Q costs $5$ rupees. Write the cost of $3$ kg of P and $4$ kg of Q.

5. Two corners cost the same. What does that tell you?