If Z=10x+25y subject to 0 leq x leq 3, 0 leq | vedclass

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(C) The constraints are $x \leq 3, y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$.We identify the vertices of the feasible region by finding the intersecti

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$(2,3)$

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(C) The constraints are $x \leq 3, y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$.We identify the vertices of the feasible region by finding the intersection points of the boundary lines:$1$. $x=0, y=0 \Rightarrow O(0,0)$$2$. $x=3, y=0 \Rightarrow A(3,0)$$3$. $x=3, x+y=5 \Rightarrow P(3,2)$$4$. $x+y=5, y=3 \Rightarrow Q(2,3)$$5$. $x=0, y=3 \Rightarrow D(0,3)$Now,we evaluate the objective function $Z=10x+25y$ at each vertex:- At $O(0,0): Z = 10(0) + 25(0) = 0$- At $A(3,0): Z = 10(3) + 25(0) = 30$- At $P(3,2): Z = 10(3) + 25(2) = 30 + 50 = 80$- At $Q(2,3): Z = 10(2) + 25(3) = 20 + 75 = 95$- At $D(0,3): Z = 10(0) + 25(3) = 0 + 75 = 75$The maximum value of $Z$ is $95$,which occurs at the point $(2,3)$.Therefore,the correct option is $C$.

If $Z=10x+25y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$,then $Z$ is maximum at the point:

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