Math Problem Statement

YOu have $32,000 you would like to invest in two different stocks, Stock Boll and Stock Coff. The buying limit on Stock Coff is $12,600 and you want to spend at least four times as much money on Stock Boll as Stock Coff. If Stock Boll earns 6% annual interest and Stock Coff earns 4% annual interest, how much money (in dollars) should you invest in each stock to maximize your annual interest earned? What is the maximum annual interest (in dollars)? You will earn a maximum total annual interest of $

Solution

Let's define variables to represent the amounts of money invested in each stock:

  • Let xx be the amount invested in Stock Coff.
  • Let yy be the amount invested in Stock Boll.

We know the following:

  1. The total investment is $32,000, so: x+y=32,000x + y = 32,000
  2. The buying limit for Stock Coff is $12,600, so: x12,600x \leq 12,600
  3. You want to invest at least four times as much in Stock Boll as in Stock Coff, so: y4xy \geq 4x
  4. Stock Coff earns 4% annual interest and Stock Boll earns 6% annual interest. The total interest II earned is given by: I=0.04x+0.06yI = 0.04x + 0.06y

Step 1: Express yy in terms of xx

From the first equation x+y=32,000x + y = 32,000, we can express yy as: y=32,000xy = 32,000 - x

Step 2: Substitute yy into the inequality y4xy \geq 4x

Substituting y=32,000xy = 32,000 - x into y4xy \geq 4x gives: 32,000x4x32,000 - x \geq 4x Solving this inequality: 32,0005x32,000 \geq 5x x32,0005=6,400x \leq \frac{32,000}{5} = 6,400

Step 3: Determine the feasible range for xx

We have two constraints on xx:

  • From the buying limit: x12,600x \leq 12,600
  • From the inequality: x6,400x \leq 6,400

Thus, the maximum amount we can invest in Stock Coff is x=6,400x = 6,400. The feasible range for xx is: 0x6,4000 \leq x \leq 6,400

Step 4: Calculate the corresponding value of yy

When x=6,400x = 6,400, we substitute this into y=32,000xy = 32,000 - x to get: y=32,0006,400=25,600y = 32,000 - 6,400 = 25,600

Step 5: Calculate the total interest earned

Now, we compute the total interest earned using x=6,400x = 6,400 and y=25,600y = 25,600: I=0.04(6,400)+0.06(25,600)I = 0.04(6,400) + 0.06(25,600) I=256+1,536=1,792I = 256 + 1,536 = 1,792

Final Answer:

  • Amount to invest in Stock Coff: $6,400
  • Amount to invest in Stock Boll: $25,600
  • Maximum annual interest: $1,792

Do you need further details or clarifications? Here are five related questions to explore:

  1. What happens to the total interest if the interest rates for both stocks are switched?
  2. How would the solution change if you wanted to invest at least 3 times as much in Stock Boll instead of 4 times?
  3. What is the effect on the maximum interest if the total amount to invest is increased to $40,000?
  4. How much less annual interest would you earn if the buying limit on Stock Coff were lowered to $10,000?
  5. What would be the total interest if Stock Boll's interest rate increased to 8%?

Tip: Always check your constraints carefully in optimization problems to ensure you're maximizing your return while staying within limits.

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Math Problem Analysis

Mathematical Concepts

Linear Inequalities
Optimization
Interest Calculation

Formulas

x + y = 32,000
y ≥ 4x
I = 0.04x + 0.06y

Theorems

Linear Programming

Suitable Grade Level

Grades 10-12