Determine the values of x and y that produce the minimum value of the objective function on the feasible
region. Determine the minimum value of the objective function on the feasible region.
-$$\begin{array} { l }
x \geq 0 , y \geq 0 \
40 x + 30 y \geq 3,200 \
x \geq 1.75 y \
\quad \text { Minimize } z = 22 x + 12 y
\end{array}$$
A) $x = 32 , y = 56$; Minimum value: 48,320
B) $x = 32 , y = 56$; Minimum value: 1,376
C) $x = 56 , y = 32$; Minimum value: 60,800
D) $x = 56 , y = 32$; Minimum value: 1,616

Question

Determine the values of x and y that produce the minimum value of the objective function on the feasible
region. Determine the minimum value of the objective function on the feasible region.
-$$\begin{array} { l } 
x \geq 0 , y \geq 0 \
40 x + 30 y \geq 3,200 \
x \geq 1.75 y \
\quad \text { Minimize } z = 22 x + 12 y
\end{array}$$
A) $x = 32 , y = 56$; Minimum value: 48,320
B) $x = 32 , y = 56$; Minimum value: 1,376
C) $x = 56 , y = 32$; Minimum value: 60,800
D) $x = 56 , y = 32$; Minimum value: 1,616

Quizplus · Accepted Answer

To find the minimum value of the objective function $z = 22x + 12y$ given the constraints, we first identify the feasible region defined by the constraints $x \geq 0$, $y \geq 0$, $40x + 30y \geq 3200$, and $x \geq 1.75y$. The minimum value of $z$ occurs at a vertex of the feasible region. By substituting the equality conditions of the constraints into the objective function, we can find the values of $x$ and $y$ that minimize $z$. Given the constraints, the point $x = 56, y = 32$ is on the boundary of the feasible region and satisfies all constraints. Substituting these values into the objective function gives $z = 22(56) + 12(32) = 1232 + 384 = 1616$, which is the minimum value of the objective function.

Determine the Values of X and Y That Produce the Minimum