Absolute maximum and absolute minimum value of $f(x,y)=20-8x+12y$ over the closed triangular region formed by the points $(0,0),(12 ,0),(12,16).$
What i try: firstci calculate partial derivative
$f_{x}(x,y)=-8$ and $f_{y}(x,y)=12$
Here partial derivativecare constant.
Means there is no critical point for function.
Now we will calculate the value of $f(x,y)$ in closed region
$\bullet$ Along $x$ axis (where $0\leq x\leq 12$)
$f(x,0)=20-8x$ for $x\in[0,12]$
So for maximum , put $x=0$ we have $f(0,0)=20$
For minimum , put $x=12$ we have $f(12,0)=-76$
$\bullet$ along parallel to $y$ i.e alnog $x=12$ line axis
$f(12,y)=12y-76$ where $0\leq y\leq 16$
For maximum , put $y=16$ we have $f(12,16)=12\cdot 16-76=$
For minimum , put $y=0$ we have $f(12,0)=-76$
Now we will calculate along line $y=(4/3)x$
Put int $f(x,y)=20-8x+12y=20+8x$
How do i solve it after that Help me please , thanks
Best Answer
Let $f(x,y)=k$.
Thus, $$y=\frac{2}{3}x+\frac{k}{12}-\frac{5}{3}.$$
Since, the maximum and the minimum occur on the vertex of the domain and the line $y-6=\frac{2}{3}(x-12)$ is placed "inside the domain", we see that the maximum and the minimum $f$ gets in the points $(0,0)$ and $12,0)$.
For $(0,0)$ we obtain: $$0=\frac{k}{12}-\frac{5}{3},$$ which gives $$k=20.$$ For $(12,0)$ we obtain: $$0=8+\frac{k}{12}-\frac{5}{3},$$ which gives $$k=-76.$$
We got $$\max{f}=20$$ and $$\min{f}=-76.$$