College Algebra Applications: Geometry, Percentages, Simple Interest, and Mixture Problems

Geometry Applications: The Isosceles Triangle Problem

Vietnam Veterans Memorial Problem

This problem involves finding the lengths of the two equal sides of an isosceles triangle, given the base and the perimeter. Such problems are common in College Algebra, where geometric relationships are solved using algebraic equations.

• Isosceles Triangle: A triangle with two sides of equal length.

• Given: Base = 438 ft, Perimeter = 931.5 ft.

• Let the length of each equal side be $x$.

Equation Setup:

• Perimeter formula: $438 + 2x = 931.5$

• Solve for $x$:

$2x = 931.5 - 438$ $2x = 493.5$ $x = \frac{493.5}{2} = 246.75$

• Answer: Each wall is 246.75 ft long.

Example Application: This type of problem is useful in architecture and design, where dimensions must be calculated based on constraints.

Percentages and Commissions

Real Estate Commission Problem

Problems involving commissions require understanding how to calculate a percentage of a total amount and subtract it to find the net proceeds.

• Given: Sale price = $159,000, Commission rate = 6%.

• Commission Amount: $159,000 \times 0.06 = $9,540$

• Net Amount Received: $159,000 - $9,540 = $149,460$

Example Application: Calculating commissions is common in real estate, sales, and finance.

Simple Interest Problems

Simple Interest Formula

Simple interest is calculated using the formula:

$I = Prt$

• $I$ = interest earned

• $P$ = principal (initial amount invested)

• $r$ = annual interest rate (as a decimal)

• $t$ = time in years

Example 1: Two-Part Investment Problem

• Given: Total interest = $1020, $1000 less invested at 4.5% than at 3%, time = 1 year.

• Let $x$ = amount invested at 3%, $x - 1000$ = amount invested at 4.5%.

• Equation: $0.03x + 0.045(x - 1000) = 1020$

• Solve for $x$:

$0.03x + 0.045x - 45 = 1020$ $0.075x = 1065$ $x = \frac{1065}{0.075} = 14,200$ $\text{At 3%: } 14,200$ $\text{At 4.5%: } 14,200 - 1000 = 13,200$

• Answer: $14,200 at 3% and $13,200 at 4.5%.

Example 2: Another Two-Part Investment Problem

• Given: $5000 more invested at 3.5% than at 4%, total interest = $1440, time = 1 year.

• Let $x$ = amount invested at 4%, $x + 5000$ = amount invested at 3.5%.

• Equation: $0.04x + 0.035(x + 5000) = 1440$

• Solve for $x$:

$0.04x + 0.035x + 175 = 1440$ $0.075x = 1265$ $x = \frac{1265}{0.075} = 16,866.67$ $\text{At 4%: } 16,866.67$ $\text{At 3.5%: } 16,866.67 + 5000 = 21,866.67$

• Answer: $16,866.67 at 4% and $21,866.67 at 3.5%.

Mixture Problems

Mixing Solutions with Different Concentrations

Mixture problems involve combining two or more solutions of different concentrations to obtain a solution of a desired concentration. These are solved using systems of equations.

Example 1: Mixing Alcohol Solutions

• Given: 12 L of 12% solution, $x$ L of 20% solution, final mixture is 14% solution.

• Equation: $0.12 \times 12 + 0.20x = 0.14(12 + x)$

• Solve for $x$:

$1.44 + 0.20x = 1.68 + 0.14x$ $0.20x - 0.14x = 1.68 - 1.44$ $0.06x = 0.24$ $x = \frac{0.24}{0.06} = 4$

• Answer: 4 L of 20% solution are needed.

Example 2: Mixing to Obtain a Higher Concentration

• Given: $x$ L of 10% solution, 40 L of 50% solution, final mixture is 40% solution.

• Equation: $0.10x + 0.50 \times 40 = 0.40(x + 40)$

• Solve for $x$:

$0.10x + 20 = 0.40x + 16$ $20 - 16 = 0.40x - 0.10x$ $4 = 0.30x$ $x = \frac{4}{0.30} \approx 13.33$

• Answer: Approximately 13.33 L of 10% solution are needed.

Summary Table: Types of Algebra Word Problems

Problem Type	Key Formula/Approach	Example
Geometry (Perimeter)	$\text{Perimeter} = \text{sum of all sides}$	Find lengths of triangle sides
Commission	$\text{Commission} = \text{Rate} \times \text{Total}$	Real estate sale
Simple Interest	$I = Prt$	Investment at different rates
Mixture	$\text{Amount}_1 \times \text{Conc}_1 + \text{Amount}_2 \times \text{Conc}_2 = \text{Total Amount} \times \text{Final Conc}$	Mixing solutions

Additional info: The above notes expand on the original problems by providing step-by-step solutions, general formulas, and a summary table for quick reference. These are standard applications in College Algebra courses.