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
$\endgroup$ 12 Answers
$\begingroup$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 $$\min{f}=-76.$$
$\endgroup$ 1 $\begingroup$The approach used by Michael Rosenberg looks at the line intersection of the "level surfaces" $ \ f(x,y) \ = \ 20 \ - \ 8x \ + \ 12y \ \ = \ \ k \ $ with the specified triangle. Another way to examine this is to consider "slices" through that triangle parallel to one of the coordinate axes; here, we will use slices $ \ y \ = \ Y \ $ and investigate the behavior of the function along them. (Using "vertical slices" $ \ x \ = \ X \ $ will lead to a similar discussion.)
For any level surface of $ \ f \ $ , the plane described inclines "downward" with increasing $ \ x \ $ and upward with increasing $ \ y \ $ , so we might reasonably expect that the values of $ \ f \ $ are smaller near the vertex $ \ (12,0) \ $ and larger near the other vertices. Along any slice $ \ y \ = \ Y \ $ , our function becomes $ \ f(x,Y) \ = \ (20 \ + \ 12Y) \ - \ 8x \ $ ; as this is a linear function, there is no relative extremum along the slice, so we only need to examine the endpoints.
The side connecting the origin to $ \ (12,16) \ $ lies on the line $ \ y \ = \ \frac{4}{3} \ x \ $ . so each of our slices covers the interval $ \left[ \ \frac{3}{4} Y \ , \ 12 \ \right] $ for $ \ x \ . $ The values of $ \ f \ $ at the endpoints are then
$$ f \left(\frac{3}{4} Y \ , \ Y \right) \ = \ \ (20 \ + \ 12Y) \ - \ 8·\frac{3}{4} Y \ \ = \ \ 20 \ + \ 6Y \ \ , $$
$$ f \left(12 \ , \ Y \right) \ = \ \ (20 \ + \ 12Y) \ - \ 8·12 \ \ = \ \ 12Y \ - \ 76 \ \ . $$
We see at once that $ f (0 \ , \ 0) \ = \ 20 \ $ and $ \ f (12 \ , \ 0) \ = \ -76 \ \ . $
We could then either use the coordinates of the third vertex, or simply note that the ends of the slices meet where the function values are equal; this suffices because both functions are increasing with increasing $ \ Y \ . $ Thus,
$$ f \left(\frac{3}{4} Y \ , \ Y \right) \ = \ \ f \left(12 \ , \ Y \right) \ \ \ \Rightarrow \ \ \ 20 \ + \ 6Y \ \ = \ \ 12Y \ - \ 76 $$
$$ \Rightarrow \ \ \ 96 \ \ = \ \ 6Y \ \ \ \Rightarrow \ \ \ Y \ = \ 16 \ \ \ . $$
For the third vertex, we have
$$ f (12 \ , \ 16 ) \ \ = \ \ 12·16 \ - \ 76 \ \ = \ \ 116 \ \ . $$
So the absolute minimum for the triangular region occurs at $ \ f (12 \ , \ 0) \ = \ -76 \ $ and the absolute maximum at $ \ f (12 \ , \ 16) \ = \ 116 \ \ . $
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