Question

A biologist is developing two new strains of bacteria. Each sample of Type 1 bacteria produces four new viable bacteria, and each sample of Type II bacteria produces three new viable bacteria. Altogether, at least 240 new viable bacteria must be produced. At least 20, but not more than 60, of the original samples must by Type I. Not more than 70 of the original samples can be Type II. A sample of Type I costs $5 and a sample of Type II costs $7. How many samples of each bacteria should the biologist use to minimize the cost? What is the minimum cost?

Answer 1

The cost can be optimized by using a Linear Programming given the  linear constraint system

• To minimize the cost, the biologist should use 60 samples of Type I bacteria and 0 samples of Type II bacteria

Reason:

Let X represent Type 1 bacteria, and let Y, represent Type II bacteria, we have;

The constraints are;

4·X + 3·Y ≥ 240

20 ≤ X ≤ 60

Y ≤ 70

P = 5·X + 7·Y

Solving the inequality gives;

4·X + 3·Y ≥ 240

• $Y \geq 80 - \dfrac{4}{3} \cdot X$ (Equation for the inequality graphs)

The boundary of the feasible region are;

(20, 70)

(20, 53.$\overline 3$)

(60, 0)

(60, 70)

The cost are ;

$\begin{array}{|c|c|c|}X&Y&P= 5\times X + 7 \times Y\\20&70&590\\20&53.\overline 3&473.\overline 3\\60&0&300\\60&70&790\end{array}\right]$

• Therefore, the minimum cost of $300 is obtained by using 60 samples of Type I and 0 samples of Type II

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