Quadratic Functions Word Problems 1 worksheet with math exercises and a diagram.

Problem: Quadratic Functions Word Problems

#### Part 1: Revenue Maximization
The demand function for a product is given by:
\[ p = -\frac{1}{2}x + 400 \]
where \( p \) is the unit price and \( x \) is the number of units sold. The revenue \( R \) is given by:
\[ R = px \]

We need to determine the unit price \( p \) that maximizes the revenue.

##### Step 1: Express Revenue as a Function of \( x \)
The revenue \( R \) is the product of the unit price \( p \) and the number of units sold \( x \):
\[ R = p \cdot x \]
Substitute the demand function \( p = -\frac{1}{2}x + 400 \) into the revenue equation:
\[ R = \left( -\frac{1}{2}x + 400 \right)x \]
\[ R = -\frac{1}{2}x^2 + 400x \]

##### Step 2: Find the Maximum Revenue
The revenue function \( R(x) = -\frac{1}{2}x^2 + 400x \) is a quadratic function in the standard form \( R(x) = ax^2 + bx + c \), where:
\[ a = -\frac{1}{2}, \quad b = 400, \quad c = 0 \]

Since \( a < 0 \), the parabola opens downward, and the maximum value occurs at the vertex. The \( x \)-coordinate of the vertex of a parabola is given by:
\[ x = -\frac{b}{2a} \]
Substitute \( a = -\frac{1}{2} \) and \( b = 400 \):
\[ x = -\frac{400}{2 \left( -\frac{1}{2} \right)} \]
\[ x = -\frac{400}{-1} \]
\[ x = 400 \]

##### Step 3: Find the Corresponding Unit Price
Using the demand function \( p = -\frac{1}{2}x + 400 \), substitute \( x = 400 \):
\[ p = -\frac{1}{2}(400) + 400 \]
\[ p = -200 + 400 \]
\[ p = 200 \]

##### Final Answer for Part 1:
The unit price that maximizes the revenue is:
\[ \boxed{200} \]

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#### Part 2: Largest Rectangle Inscribed in a Right Triangle
A rectangle is inscribed in a right triangle with legs of 3 meters and 4 meters. The perimeter of the rectangle is 20 feet. We need to find the dimensions of the rectangle with the largest area.

##### Step 1: Set Up the Coordinate System
Place the right triangle in the coordinate plane with:
- The right angle at the origin \((0, 0)\),
- One leg along the \( x \)-axis (length 4 meters),
- The other leg along the \( y \)-axis (length 3 meters).

The hypotenuse of the triangle has the equation:
\[ \frac{x}{4} + \frac{y}{3} = 1 \]
or equivalently:
\[ y = 3 - \frac{3}{4}x \]

##### Step 2: Dimensions of the Rectangle
Let the rectangle have:
- Width \( w \) (parallel to the \( x \)-axis),
- Height \( h \) (parallel to the \( y \)-axis).

The top-right corner of the rectangle lies on the hypotenuse. If the bottom-left corner of the rectangle is at \((0, 0)\), then the coordinates of the top-right corner are \((w, h)\). Since this point lies on the hypotenuse, it satisfies the equation of the hypotenuse:
\[ h = 3 - \frac{3}{4}w \]

##### Step 3: Perimeter Constraint
The perimeter of the rectangle is given as 20 feet. Convert 20 feet to meters (since the triangle dimensions are in meters):
\[ 1 \text{ foot} = 0.3048 \text{ meters} \]
\[ 20 \text{ feet} = 20 \times 0.3048 = 6.096 \text{ meters} \]

The perimeter of the rectangle is:
\[ 2(w + h) = 6.096 \]
\[ w + h = 3.048 \]

Substitute \( h = 3 - \frac{3}{4}w \) into the perimeter equation:
\[ w + \left( 3 - \frac{3}{4}w \right) = 3.048 \]
\[ w + 3 - \frac{3}{4}w = 3.048 \]
\[ \frac{1}{4}w + 3 = 3.048 \]
\[ \frac{1}{4}w = 0.048 \]
\[ w = 0.192 \]

Now, solve for \( h \):
\[ h = 3 - \frac{3}{4}(0.192) \]
\[ h = 3 - 0.144 \]
\[ h = 2.856 \]

##### Step 4: Area of the Rectangle
The area \( A \) of the rectangle is:
\[ A = w \cdot h \]
\[ A = 0.192 \cdot 2.856 \]
\[ A \approx 0.548 \text{ square meters} \]

##### Final Answer for Part 2:
The dimensions of the rectangle with the largest area are approximately:
\[ \boxed{0.192 \text{ meters by } 2.856 \text{ meters}} \]

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#### Part 3: Cookie Sales and Revenue
Nate opened a cookie store and found that the relationship between the price of a cookie \( p \) (in dollars) and the number of cookies sold \( n \) is given by:
\[ n = 500 - 100p \]

He also knows that the cost to make each cookie is $0.50. He wants to maximize his profit. The profit \( P \) is given by:
\[ P = \text{Revenue} - \text{Cost} \]
\[ \text{Revenue} = p \cdot n \]
\[ \text{Cost} = 0.50 \cdot n \]

##### Step 1: Express Profit as a Function of \( p \)
The number of cookies sold is:
\[ n = 500 - 100p \]

The revenue is:
\[ \text{Revenue} = p \cdot n = p(500 - 100p) = 500p - 100p^2 \]

The cost is:
\[ \text{Cost} = 0.50 \cdot n = 0.50(500 - 100p) = 250 - 50p \]

The profit is:
\[ P = \text{Revenue} - \text{Cost} \]
\[ P = (500p - 100p^2) - (250 - 50p) \]
\[ P = 500p - 100p^2 - 250 + 50p \]
\[ P = -100p^2 + 550p - 250 \]

##### Step 2: Maximize the Profit Function
The profit function \( P(p) = -100p^2 + 550p - 250 \) is a quadratic function in the standard form \( P(p) = ap^2 + bp + c \), where:
\[ a = -100, \quad b = 550, \quad c = -250 \]

The maximum profit occurs at the vertex of the parabola. The \( p \)-coordinate of the vertex is:
\[ p = -\frac{b}{2a} \]
\[ p = -\frac{550}{2(-100)} \]
\[ p = \frac{550}{200} \]
\[ p = 2.75 \]

##### Final Answer for Part 3:
The price Nate should set for the cookies to maximize his profit is:
\[ \boxed{2.75} \]

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Final Answers:

1. The unit price that maximizes revenue is: \( \boxed{200} \)
2. The dimensions of the rectangle with the largest area are: \( \boxed{0.192 \text{ meters by } 2.856 \text{ meters}} \)
3. The price Nate should set for the cookies is: \( \boxed{2.75} \)