Find the absolute extrema of the function $f ( x , y ) = 2 x + 3 y - 5$ on the closed triangular region with vertices $( 0,0 )$ , $( 5,0 )$ and $( 5,4 )$ .
A) Absolute minimum 0, Absolute maximum 5
B) Absolute minimum -5, Absolute maximum 5
C) Absolute minimum -5, Absolute maximum 17
D) Absolute minimum 5, Absolute maximum 17

Question

Quizplus · Accepted Answer

First, we find the critical points of $f(x,y)$ by setting its partial derivatives to zero:
$$ \frac{\partial}{\partial x} f(x,y) = 2 \quad 	ext{and} \quad  \frac{\partial}{\partial y} f(x,y) = 3 $$
Thus, the only critical point is $(0,0)$.

Next, we evaluate $f(x,y)$ at the vertices of the triangular region:
\begin{align*}
f(0,0) &= -5 \
f(5,0) &= 5 \
f(5,4) &= 17
\end{align*}

Finally, we evaluate $f(x,y)$ on the segments connecting the vertices of the triangle. On the segment connecting $(0,0)$ to $(5,0)$, we have $y=0$ and $0\leq x \leq 5$, so $f(x,y) = 2x - 5$. The minimum value of this function is achieved at $x=0$, where $f(0,0)=-5$. The maximum value of this function is achieved at $x=5$, where $f(5,0) = 5$. On the segment connecting $(5,0)$ to $(5,4)$, we have $x=5$ and $0\leq y\leq 4$, so $f(x,y) = 3y-5$. The minimum value of this function is achieved at $y=0$, where $f(5,0) = 5$. The maximum value of this function is achieved at $y=4$, where $f(5,4)= 17$. Finally, on the segment connecting $(5,4)$ to $(0,0)$, we have $y=4-x$ and $0\leq x\leq 5$, so $f(x,y) = 2x+3(4-x)-5$. This simplifies to $f(x,y) = -x+7$, which is a decreasing function of $x$ on this segment. Thus, the maximum value of $f(x,y)$ on this segment is achieved at $x=0$, where $f(0,4) = 7$.

Comparing all the values we have found, we see that the absolute minimum of $f(x,y)$ on the triangular region is $-5$ and the absolute maximum is $17$. Therefore, the answer is $\boxed{	extbf{(C)}}$.

Find the Absolute Extrema of the Function f(x,y)=2x+3y−5f ( x , y ) = 2 x + 3 y - 5f(x,y)=2x+3y−5

Find the Absolute Extrema of the Function $f ( x , y ) = 2 x + 3 y - 5$