Answer to Question #186095 in Math for Momin ali

Question

Solve the following

𝑎) Maximize 𝑓(𝑥, 𝑦) = 2𝑥 + 5𝑦; subject to the constraints.

2𝑦 − 𝑥 ≤ 8; 𝑥 − 𝑦 ≤ 4; 𝑥 ≥ 0; 𝑦 ≥ 0.

𝑏) Minimize 𝑧 = 3𝑥 + 𝑦; subject to the constraints.

3𝑥 + 5𝑦 ≥ 15; 𝑥 + 6𝑦 ≥ 9; 𝑥 ≥ 0; 𝑦 ≥ 0.

Answer 1

a)

$f(x, y)=2x+5y$

$OA: x=0, 0\leq y\leq4$

$$f(0, y)=5y$$

$$f(0, 0)=0, f(0, 4)=20$$

$AB: 2y-x=8=>x=2y-8, 4\leq y\leq12$

$$f(2y-8, y)=2(2y-8)+5y=9y-16$$

$$f(0, 4)=20, f(16, 12)=92$$

$BC: x-y=4=>x=y+4, 0\leq y\leq12$

$$f(y+4, y)=2(y+4)+5y=7y+8$$

$$f(16, 12)=92, f(4, 0)=8$$

$CO: y=0, 0\leq x\leq4$

$$f(x, 0)=2x$$

$$f(0, 0)=0, f(4,0)=8$$

The function $f(x, y)$ has maximum with value of $92$ at $x=16, y=12$

subject to the constraints.

b)

$z=3x+y$

$$\begin{matrix} 3x+5y=15 \\ x+6y=9 \end{matrix}$$

$x=\dfrac{45}{13}, y=\dfrac{12}{13}$

$x=0, y\geq3$

$$z=y\geq3$$

$$z(0, 3)=3$$

$AB: 3x+5y=15=>x=5-\dfrac{5}{3}y, \dfrac{12}{13}\leq y\leq3$

$$z=15-5y+y=15-4y$$

$$z(0, 3)=3, z(\dfrac{45}{13}, \dfrac{12}{13})=\dfrac{147}{13}$$

$BC: x+6y=9=>x=9-6y, 0\leq y\leq\dfrac{12}{13}$

$$z=27-18y+y=27-17y$$

$$z(\dfrac{45}{13}, \dfrac{12}{13})=\dfrac{147}{13}, z(9, 0)=27$$

$y=0, x\geq9$

$$z=3x\geq27$$

The function $z$ has minimum with value of $3$ at $x=0, y=3$

subject to the constraints.