Answer to Question #248092 in Math for Silas Petrus

Question

Solve the following linear programming graphically [5] Minimize: 𝑧 = 200𝑥 + 500𝑦 Subject to the constraints: 𝑥 + 2𝑦 ≥ 10 3𝑥 + 4𝑦 ≤ 24 𝑥 ≥ 0; 𝑦 ≥ 

Answer 1

Minimize $𝑧 = 200𝑥 + 500𝑦$

subject to the constraints

$$\begin{matrix} x+2y\geq10 \\ 3𝑥 + 4𝑦 \leq 24 \\ x\geq0, y\geq0 \end{matrix}$$

Find the point(s) of intersection

$$y=-\dfrac{1}{2}x+5$$

$$y=-\dfrac{3}{4}x+6$$

$x=0:$

$$y=-\dfrac{3}{4}(0)+6, Point\ A(0,6)$$

$$y=-\dfrac{1}{2}(0)+5, Point\ C(0,5)$$

$$-\dfrac{1}{2}x+5=-\dfrac{3}{4}x+6$$

$$\dfrac{1}{4}x=1$$

$$x=4, y=3,Point\ B(4,3)$$

Point $A(0,6):𝑧(0,6) = 200(0) + 500(6)=3000$

Point $B(4,3):𝑧(4,3) = 200(4) + 500(3)=2300$

Point $C(05):𝑧(0,5) = 200(0) + 500(5)=2500$

The function $z$ has a minimum with value of $2300$ at $(4,3).$