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Question 1080575: On your birthday your great aunt gave you $17,000
. You would like to invest at least $8500 of the money in municipal bonds yielding 4% and no more than $3500 in Treasury bills yielding 6%. How much should be placed in each investment in order to maximize the interest earned in one year? Assume simple interest applies. Let x represent the amount of money in municipal bonds and y
represent the amount of money in Treasury bills.
Write the objective function, f(x,y)
, to represent the interest earned in one year.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you want to invest at least 8500 in municipal bonds yielding 4%
you want to invest at most 3500 in treasury bills yielding 6%
without doing anything else, it appears that maximizing the amount invested at 6% would yield the most interest total.
therefore:
3500 * .06 = 210
17000 - 3500 = 13500 * .04 = 540
total interest is equal to 750.
that's the maximum interest that can be earned from 17000 given the constraints.
treating this a linear entineering type problem, you would do the following:
x = amount invested at 4%
y = amount invested at 6%.
constraints are:
x >= 8500
y <= 3500
x + y <= 17000
x >= 0
y >= 0
x + y <= 17000 is used instead of x + y = 17000 because it was not specifically stipulated that you had to invest all 17000, although it's peretty clear that, unless you do that, you won't be able to maximize your interest as will be seen from the corner points of the graph i will show you below:
your objective function is interest = .04x + .06y
that's what you want to maximize.
using the graphing calculator at www.desmos.com, you would graph the opposite of the inequalities.
doing this, your region of feasibility will be the area of the graph that is not shaded.
your graph will look like this:
you would analyze your objective function at the corner points of the feasible region.
once you do that, you will see that the maximum interest is earned when x = 13,500 anf y = 3500.
that interest earned is $750 in one year.
all the constraints are met.
x >= 8500
y <= 3500
x + y <= 17000
x >= 0
y >= 0
if you look at the equations shown on the graph, you will see that the opposite of these inequalities is what was graphed.
that's what makes the unshaded area of the graph equal to the feasible region.
doing it this way makes the feasible region easier to see.
you can see what i mean if you graph the inequalities themselves rather than the reverse of the inequalities.
the feasible region will be shaded but so will other areas of the graph.
you would need to find the shaded region that was most dark in order to locate the feasible region. it can be done but it's more difficult to see.
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