you can solve it graphically.
you have unfortunately not specified whether you want AT LEAST 1/3 of the total portfolio to be allocated to type A investments, or AT MOST 1/3 of the total portfolio to be allocated to type A investments.
more unfortunately, it appears your selections include results that are applicable to both scenarios.
i'll solve the problems for you both ways and you can pick the answer that fits the problem.
option 1 will assume you meant AT LEAST in reference to type A investment.
option 2 will assume you meant AT MOST in reference to type A investment.
the graphing calculator i used can be found at the following link:
https://www.desmos.com/calculator
the method i chose is based on the use of this calculator.
if you don't use this particular calculator, then your method may be different, but the concepts of how to solve these problems graphically are the same.
option 1 of your first problem is shown below:
1. An investor has $600,000 to invest in two types of investments. Type A pays 7% annually and type B pays 9% annually. To have a well-balanced portfolio, the investor imposes the following conditions. AT LEAST 1/3 of the total portfolio is to be allocated to type A investments and at least 1/3 of the portfolio is to be allocated to the type B investments. What is the optimal amount that should be invested in each investment?
A. $210,000 in type A (7%); $390,000 in type B (9%)
B. $400,000 in type A (7%); $200,000 in type B (9%)
C. $200,000 in type A (7%); $400,000 in type B (9%)
D. $600,000 in type A (7%); $0 in type B (9%)
E. $0 in type A (7%); $600,000 in type B (9%)
let x equal the amount to be invested in type A investment.
let y equal the amount to be invested in type B investment.
your objective function is interest = .07x + .09y
your constraint functions are:
x >= 0
y >= 0
x + y <= 600,000
x >= 1/3 * (x + y) ***** for option 1 only
y >= 1/3 * (x + y)
you graph the opposite of these inequalities.
you will graph:
x <= 0
y <= 0
x + y >= 600,000
x <= 1/3 * (x + y) ***** for option 1 only
y <= 1/3 * (x + y)
your region of feasibility will be the area of the graph that is not shaded.
your optimal solution will be at the corner points.
this is your graph for option 1 of problem 1:
your corner points are:
x = 200,000 and y = 400,000
x = 400,000 and y = 200,000
your analyze your objective function of interest = .07x + .09y at these corner points.
200,000 * .07 + 400,000 * .09 gives you interest of 50,000
400,000 * .07 +200,000 * .09 give you interest of 46,000
your optimal solution is when you x = 200,000 and y = 400,000
this means 200,000 is invested in type A investment and 400,000 is invested in type B solution.
that would be selection C.
if your first problem was mean to be as follows instead:
1. An investor has $600,000 to invest in two types of investments. Type A pays 7% annually and type B pays 9% annually. To have a well-balanced portfolio, the investor imposes the following conditions. AT MOST 1/3 of the total portfolio is to be allocated to type A investments and at least 1/3 of the portfolio is to be allocated to the type B investments. What is the optimal amount that should be invested in each investment?
A. $210,000 in type A (7%); $390,000 in type B (9%)
B. $400,000 in type A (7%); $200,000 in type B (9%)
C. $200,000 in type A (7%); $400,000 in type B (9%)
D. $600,000 in type A (7%); $0 in type B (9%)
E. $0 in type A (7%); $600,000 in type B (9%)
that would be option 2.
your objective function would remain the same at .07x + .09y
your constraint functions would be:
x >= 0
y >= 0
x + y <= 600,000
x <= 1/3 * (x + y) ***** this is for option 2 only
y >= 1/3 * (x + y)
you graph the opposite of these inequalities.
you will graph:
x <= 0
y <= 0
x + y >= 600,000
x >= 1/3 * (x + y) ***** this is for option 2 only
y <= 1/3 * (x + y)
your region of feasibility will be the area of the graph that is not shaded.
your optimal solution will be at the corner points.
this is your graph for option 2 of problem 1:
your corner points are:
x = 0 and y = 600,000
x = 200,000 and y = 400,000
you analyze your objective function of interest = .07x + .09y at these corner points.
0 * .07 + 600,000 * .09 gives you interest of 54,000
200,000 * .07 + 400,000 * .09 gives you interest of 50,000
your optimal solution is when x = 0 and y = 600,000
this means 0 is invested in type A investment and 600,000 is invested in type B solution.
that would be selection E.
it obviously makes a difference whether you meant at least or at most for the type A investment.
once you determine that, you can find the solution that is applicable.
i'll give you the objective constraint functions for your second problem along with the graphs, but without all the extra verbiage.
if you understood what i did for the first problem, then you should be able to figure out the proper solution for the second problem.
2. An investor has $300,000 to invest in two types of investments. Type A pays 4% annually and type B pays 6% annually. To have a well-balanced portfolio, the investor imposes the following conditions. AT ..... 1/3 of the total portfolio is to be allocated to type A investments and at least 1/3 of the portfolio is to be allocated to the type B investments. What is the optimal amount that should be invested in each investment?
A. $0 in type A (4%); $300,000 in type B (6%)
B. $110,000 in type A (4%); $190,000 in type B (6%)
C. $100,000 in type A (4%); $200,000 in type B (6%)
D. $300,000 in type A (4%); $0 in type B (6%)
E. $200,000 in type A (4%); $100,000 in type B (6%)
option 1 assumes at least 1/3 of the total portfolio is to be allocated to type A.
objective function is interest = .04 * x + .06 * y
constraint functions are:
x >= 0
y >= 0
x + y <= 300,000
x <= 1/3 * (x + y) ***** this if for option 1 only.
y <= 1/3 * (x + y)
using the www.desmos.com/calculator, you graph the opposite of these inequalities.
your graph looks like this:
at x= 100k and y = 200k, your interest is .04 * 100k + .06 * 200k = 16,000
at x = 200k and y = 100k, your interest is .04 * 200k + .06 * 100k = 14,000
your optimal solution is when 100k is invested in investment type A and 200k is invested in investment type B
that's selection C i believe.
option 2 assumes at most 1/3 of the total investment is in type A.
all objective functions and all constraint functions are the same except:
x >= 1/3 * (x + y) becomes x <= 1/3 * (x + y)
you graph the opposite of all the constraints.
your graph looks like this:
your optimal solution is when x = 0 and y = 300,000
your interest at that point is 18,000.
that looks like selection A.
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doing it the way i did using the desmos.com calculator is the most convenient and easy way to do it.
if you used other calculators, or you graphed it manually, you would do something more like the following:
objective function would be the same.
constraint functions would be the same.
you would solve for y in each of the constraint functions.
you would graph the equality portion of the inequalities.
you would then shade the area of the graph that satisfies the inequalities.
your answer would be the same.
here's a reference on linear programming type problems solved through graphing that might help you.
http://www.purplemath.com/modules/linprog.htm