# Maximum and Minimum Value Word Problems - Quadratic Equations

TLDRThis educational transcript discusses solving word problems involving quadratic functions. It first explains how to maximize a company's profit by selling 250 units, using the quadratic equation 50x - 0.1x^2 + 300, and finding a maximum profit of $6550. Next, it tackles a physics-related problem of projectile motion, using the height function h(t) = 100 + 30t - 4.9t^2, to determine the time (3.06 seconds) and height (145.92 meters) at which a ball reaches its peak after being launched from a cliff. The explanation simplifies complex concepts, making them accessible for students.

###### Takeaways

- ๐ The profit of a company can be modeled by a quadratic equation: 50x - 0.1x^2 + 300, where x is the number of units sold.
- ๐ To maximize profit, the company should find the vertex of the quadratic function, which represents the maximum point on the graph.
- ๐ The x-coordinate of the vertex can be found using the formula -b/(2a), where b is the coefficient of x and a is the coefficient of x^2.
- ๐งฎ For the given profit equation, the x-value (units to sell for maximum profit) is calculated as 250 units by using the formula with b=50 and a=-0.1.
- ๐ฐ The maximum profit is found by substituting x = 250 back into the original profit equation, resulting in a profit of $6550.
- ๐ The script also discusses projectile motion, specifically the height of a ball thrown upwards, modeled by the equation h(t) = 100 + 30t - 4.9t^2, where t is time in seconds and h is height in meters.
- ๐ The initial height (100 meters) suggests the ball is launched from a cliff or similar height.
- ๐ The initial velocity of the ball is represented by the coefficient 30, and the acceleration due to gravity is represented by 4.9 (half of 9.8 m/s^2).
- โฑ To find the time it takes for the ball to reach its maximum height, use the formula -b/(2a) with b=30 and a=-4.9, resulting in approximately 3.06 seconds.
- ๐ The maximum height is calculated by substituting t = 3.06 into the height equation, yielding a maximum height of 145.92 meters.
- ๐ The script provides a practical application of algebra in solving word problems related to profit maximization and projectile motion.

###### Q & A

### What is the equation given for modeling the company's profit?

-The equation given is 50x - 0.1x^2 + 300, where P represents the profit and x represents the number of units sold.

### How can the company maximize its profit according to the given equation?

-To maximize the profit, the company needs to find the vertex of the quadratic function, which occurs at x = 250 units.

### How do you find the x-value of the vertex in a quadratic equation?

-The x-value of the vertex can be found using the formula -b/2a, where b is the coefficient of x and a is the coefficient of x^2.

### What is the maximum profit the company can achieve, and how is it calculated?

-The maximum profit is $6550, calculated by substituting x = 250 into the profit equation.

### What is the equation for modeling the height of a ball in projectile motion?

-The equation is h(t) = 100 + 30t - 4.9t^2, where t is time in seconds and h(t) is the height in meters.

### How do you determine the time it takes for the ball to reach its maximum height?

-The time to reach the maximum height is found using the formula -b/2a, where b is 30 and a is -4.9. This gives t = 3.06 seconds.

### What is the significance of the coefficient 100 in the height equation?

-The coefficient 100 represents the initial height from which the ball is kicked, likely from a cliff or building.

### What does the coefficient 30 represent in the height equation?

-The coefficient 30 represents the initial velocity of the ball.

### What does the coefficient -4.9 represent in the height equation?

-The coefficient -4.9 is a combination of 1/2 and the acceleration due to gravity (9.8 m/s^2).

### What is the maximum height the ball reaches, and how is it calculated?

-The maximum height is 145.92 meters, calculated by substituting t = 3.061 into the height equation.

###### Outlines

##### ๐ Maximizing Profit with Quadratic Functions

This paragraph discusses a word problem related to maximizing profit for a company. The profit is modeled by the quadratic equation 50x - 0.1x^2 + 300, where x is the number of units sold. The speaker explains that to find the maximum profit, one should identify the vertex of the parabola represented by the equation. Using the formula -b/(2a), where b is the coefficient of x (50) and a is the coefficient of x^2 (-0.1), the vertex's x-coordinate is calculated to be 250 units. The speaker then demonstrates how to find the maximum profit by substituting x = 250 back into the original equation, resulting in a maximum profit of $6550.

##### ๐ Projectile Motion and Finding Maximum Height

The second paragraph explores a physics-related word problem involving projectile motion, specifically the height of a ball over time. The height is given by the function h(t) = 100 + 30t - 4.9t^2, where t is time in seconds and height in meters. The initial height is 100 meters, likely from a cliff, with an initial velocity of 30 meters per second and an acceleration due to gravity factor of 4.9 (half of 9.8 m/s^2). The speaker uses the formula -b/(2a) to find the time it takes to reach maximum height, calculating it to be approximately 3.06 seconds. To find the maximum height, the time value is substituted back into the height equation, yielding a maximum height of 145.92 meters.

###### Mindmap

###### Keywords

##### ๐กProfit

##### ๐กQuadratic Function

##### ๐กVertex

##### ๐กUnits Sold

##### ๐กProjectile Motion

##### ๐กHeight Function

##### ๐กInitial Height

##### ๐กInitial Velocity

##### ๐กAcceleration Due to Gravity

##### ๐กMaximum Height

##### ๐กAlgebra

###### Highlights

The profit of a company can be modeled by the quadratic equation 50x - 0.1x^2 + 300.

To maximize profit, find the vertex of the quadratic function, which represents the maximum point.

The vertex can be found using the formula -b/(2a) where b is the coefficient of x and a is the coefficient of x^2.

For the given profit model, the vertex occurs at x = 250 units, indicating the optimal number of units to sell for maximum profit.

The maximum profit is calculated by substituting x = 250 back into the original equation, yielding a profit of $6550.

Projectile motion is introduced as a physics topic solvable with algebraic methods.

The height of a projectile can be modeled by the function h(t) = 100 + 30t - 4.9t^2, where t is time in seconds and height in meters.

The initial height of 100 meters suggests the ball is kicked from a cliff.

The initial velocity is represented by 30, and the acceleration due to gravity is factored into the -4.9 coefficient.

To find the time to reach maximum height, apply the vertex formula to the projectile motion equation.

The time to reach maximum height is approximately 3.06 seconds.

Substitute t = 3.061 into the height equation to find the maximum height of the projectile.

The maximum height achieved is 145.92 meters.

The algebraic method simplifies the process of finding maximum values in word problems related to profit and projectile motion.

Understanding the coefficients in a quadratic equation is crucial for determining the vertex and maximum values.

Practical applications of algebra include optimizing business decisions and analyzing physical phenomena like projectile motion.

###### Transcripts

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