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December 21, 2014

December 21, 2014

Posted by **Megan** on Wednesday, March 2, 2011 at 1:44pm.

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= ?

- Math-unknown encoding -
**MathMate**, Wednesday, March 2, 2011 at 2:51pmWe cannot read the question because of unknown encoding.

Please specify character encoding used (language) or post the character before x and y in the constraints.

If it is √() you can write sqrt() instead.

Also, check if the last equation has been posted correctly.

Is it

f(x,y)=kx=y

or

f(x,y)=kx-y

?

Thank you.

- Math -
**Megan**, Wednesday, March 2, 2011 at 3:00pmSuppose 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= ?

- Math -
**MathMate**, Wednesday, March 2, 2011 at 3:44pmUse your calculator to plot three graphs, as follows:

http://img203.imageshack.us/i/1299091442.png/

This way, you have an idea what to look for.

The lines are:

(blue) f1(x)=y=3x

(red) f2(x)=y=2x

(green) f3(x)=y=x/(3x-1)

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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