Question 1171791
Let's break down this problem step-by-step to formulate it as a linear programming problem and find the solution.

**1. Define Variables**

* Let `x` be the number of prey of species A captured.
* Let `y` be the number of prey of species B captured.

**2. Formulate the Objective Function**

* The objective is to minimize the energy expenditure.
* Energy cost for species A: 7 units per prey
* Energy cost for species B: 5 units per prey
* Objective function (minimize energy): `Z = 7x + 5y`

**3. Formulate the Constraints**

* **Food A:** 6x + 3y ≥ 21
* **Food B:** 2x + 3y ≥ 12
* **Food C:** 3x + 5y ≥ 21
* **Non-negativity:** x ≥ 0, y ≥ 0 (You can't have negative prey)

**4. Solve the Linear Programming Problem**

We'll use a graphical method to find the feasible region and the optimal solution.

* **Graph the Constraints:**
    * 6x + 3y = 21 (or 2x + y = 7)
    * 2x + 3y = 12
    * 3x + 5y = 21

* **Find the Intersection Points:**
    * Intersection of 2x + y = 7 and 2x + 3y = 12:
        * Subtract the first equation from the second: 2y = 5, y = 2.5
        * Substitute y = 2.5 into 2x + y = 7: 2x + 2.5 = 7, 2x = 4.5, x = 2.25
        * (2.25, 2.5)
    * Intersection of 2x + y = 7 and 3x + 5y = 21
        * multiply 2x+y=7 by 5. 10x+5y=35
        * subtract 3x+5y=21. 7x = 14 x=2
        * 2(2)+y=7. y=3
        * (2,3)
    * Intersection of 2x + 3y = 12 and 3x + 5y = 21
        * multiply 2x+3y=12 by 3. 6x+9y=36
        * multiply 3x+5y=21 by 2. 6x+10y=42
        * subtract 6x+9y=36 from 6x+10y=42. y=6
        * 2x+3(6)=12 2x=-6 x=-3. This is not in the feasible region.

* **Find the Corner Points of the Feasible Region:**
    * (0, 7) (from 2x+y=7)
    * (6,0) (from 2x+3y=12)
    * (7,0) (from 3x+5y=21)
    * (0,4) (from 2x+3y=12)
    * (0,4.2) (from 3x+5y=21)
    * (2.25,2.5)
    * (2,3)

* **Evaluate the Objective Function at Each Corner Point:**
    * Z(0, 7) = 7(0) + 5(7) = 35
    * Z(6, 0) = 7(6) + 5(0) = 42
    * Z(7, 0) = 7(7) + 5(0) = 49
    * Z(0, 4.2) = 7(0) + 5(4.2) = 21
    * Z(2.25, 2.5) = 7(2.25) + 5(2.5) = 15.75 + 12.5 = 28.25
    * Z(2, 3) = 7(2) + 5(3) = 14 + 15 = 29

* **Determine the Minimum Value:**
    * The minimum value of Z is 28.25 at (2.25, 2.5). However, since we can't have fractional prey, we need to examine the integer points surrounding (2.25,2.5). Those points are (2,3) and (3,2).
    * Z(2,3) = 29
    * Z(3,2) = 7(3)+5(2) = 21+10=31
    * The minimum integer value is 29 at (2,3).

**Results**

* **Objective Function:** Z = 7x + 5y (minimize energy)
* **Constraints:**
    * 6x + 3y ≥ 21
    * 2x + 3y ≥ 12
    * 3x + 5y ≥ 21
    * x ≥ 0, y ≥ 0
* **Minimum Value:** 29
* **Solution:** The predator should capture 2 prey of species A and 3 prey of species B.