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Suppose that y≤3x, 2x≤y and 1/3x+1/3y≤1 together with 0≤x, 0≤y.
Determine a value of k so that the function f(x,y)=kx+y has a positive maximum value on the region at two corners.
ANS: k= ?
Use your calculator to plot three graphs, as follows:
This way, you have an idea what to look for.
The lines are:
The region that satisfies all the constraints are between the red and blue lines, and below the green curve.
To find the value of k you need to find the coordinates of the two intersection points of the (green) curve with the straight lines.
This you can obtain by solving the following equations:
f1(x)=f3(x) => 3x=x/(3x-1)....(1)
f2(x)=f3(x) => 2x=x/(3x-1)....(2)
These are quadratics and provide two roots each:
Solution to (1): (x=0 or) x=4/9, y=4/3
Solution to (2): (x=0 or) x=1/2, y=1
Therefore f(x,y)=kx+y must pass through (4/9,4/3) and (1/2,1) to satisfy the last requirement. The value of k is the slope of the line passing through these two points using the usual formula:
Slope, k = (y2-y1)/(x2-x1)
Can you take it from here?
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