Suppose that y≤3x, 2x≤y and 13x+13y≤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= ?
We 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.
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:
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?
To determine the value of k that will give the function f(x,y) = kx = y a positive maximum value on the region at two corners, we need to find the maximum value of f(x,y) subject to the given inequalities.
Let's analyze each inequality one by one and find the region in which they overlap.
1. y ≤ 3x
We can rewrite this inequality as x ≥ (1/3)y. This represents a region above the line y = 3x.
2. 2x ≤ y
This inequality represents a region below the line y = 2x.
3. 13x + 13y ≤ 1
Dividing both sides by 13, we get x + y ≤ 1/13. This represents a region below the line x + y = 1/13.
4. 0 ≤ x, 0 ≤ y
These inequalities represent the region in the first quadrant of the coordinate plane.
Now, let's find the overlapping region of these inequalities by graphing them together.
The following graph represents the overlapping region of the inequalities:
|\
| \
| \ _________ y = 3x
| \
| \
|_____\ y = 2x
|
|____________ x + y = 1/13
From the graph, we can see that the overlapping region is the triangle bound by the lines y = 3x, y = 2x, and x + y = 1/13. The vertices of this triangle represent the corners of the region.
To find the maximum value of f(x,y) = kx = y at the corners, we need to calculate the corresponding values of k at each corner.
Let's find the coordinates of the corners:
1. Intersection of y = 3x and y = 2x:
Solving these equations, we get x = 1/5 and y = 3/5.
So, the coordinates of this corner are (1/5, 3/5).
2. Intersection of y = 3x and x + y = 1/13:
Solving these equations, we get x = 1/40 and y = 3/40.
So, the coordinates of this corner are (1/40, 3/40).
3. Intersection of y = 2x and x + y = 1/13:
Solving these equations, we get x = 1/30 and y = 2/30 = 1/15.
So, the coordinates of this corner are (1/30, 1/15).
Now, let's calculate the values of k at these corners:
1. For the corner (1/5, 3/5):
Since f(x,y) = kx = y, we have k = (3/5)/(1/5) = 3.
2. For the corner (1/40, 3/40):
Similarly, k = (3/40)/(1/40) = 3.
3. For the corner (1/30, 1/15):
Here, k = (1/15)/(1/30) = 2.
Therefore, we have the values of k at each corner:
Corner (1/5, 3/5): k = 3
Corner (1/40, 3/40): k = 3
Corner (1/30, 1/15): k = 2
To determine the value of k that will give the function f(x,y) = kx = y a positive maximum value on the region at two corners, we need to select the highest value among these three values of k. In this case, the maximum value is k = 3.
Hence, the value of k that gives f(x,y) = 3x = y a positive maximum value on the region at two corners is k = 3.