This inequality problem is about investment and I am having trouble writing constraints for it.
Family has up to $130,000 to invest. They decide that they want to have at least $40,000 invested in stable bonds yielding 5.5% and that no more than $60,000 should be invested in more volatile bonds yielding 11%. How much should they invest in each type of bond to maximize income if the amount in the stable bond should not exceed the amount in the more volatile bond? What is the maximum income?
Re: Investment problem!
Evidently $60,000 in each...leaves $10,000 not invested...kinda silly...you sure there's no typo?
Re: Investment problem!
This is basically a "linear programming" problem where we attempt to maximize a linear function subject to linear constraints. And, here, since there are two values to be determined, this can be done graphically. Let "x" be the number of dollars invested in the "stable bonds" and let "y" be the number of dollars invested in the "more volatile bonds". The quantity we want to maximize is the return investment, .055x+ .11y. Notice that, for a given return level, R, that is .055x+ .11y= R, which has a straight line graph on an xy- coordinate system. Different values of R will give different different lines but with the same slope (-.055/.11) so parallel lines.
The first condition, "at least $40,000 invested in stable bonds" is . Draw a vertical line on the xy-coordinate system at x= 40000 and the "feasible area" is to the right of that line. The second condition, "no more than $60,000 should be invested in more volatile bonds" is . Draw a horizontal line at y= 60000 and the "feasible area" is below that line. Finally, we have the condition that "the amount in the stable bond should not exceed the amount in the more volatile bond" or . The line y= x is at 45 degrees and saying that means the feasible area is to the left and above that line. We can calculate that the line y= x intersects the line x= 40000 at (40000, 40000) and it intersects the line y= 60000 at (60000, 60000). The lines y= 60000 and x= 40000 intersect, of course, at (40000, 60000) so the feasible area given by all three constraints is the triangle having those vertices.
(The last condition, that the total be less than 130000 is but every point of the triangle we already have satisfies that so this is a "non-operative" restraint.)
Now, the basic concept of "linear programming" is this: the set of all possible (x, y) so that 0.55x+ .11y= R are parallel lines and we can see if we look at different parallel lines that R is increasing as we move in one direction and decreasing in the other. But the maximum value will be given as the line is just moving out of the feasible region- and that happens as it crosses through a vertex. A point of the feasible region at which R has a maximum must be a vertex! So we only need check the value of R for the three vertices of this triangle.
At (40000, 60000), R= .055(40000)+ .11(60000)= 8800.
At (40000, 40000), R= .055(40000)+ .11(40000)= 6400.
At (60000, 60000), R= .055(60000)+ .11(60000)= 9900.
The largest of those is $9900 which we get by investing 60000 dollars in each. That leaves, as Wilmer pointed out, $10000 uninvested. That happened because of the "non-operative" restraint.