2009 AP Calculus AB Free Response #3
TLDRIn this educational video, Alan from Bottle Stem Coaching tackles the 3rd free response question from the 2009 AP Calculus exam. The problem involves a company, Mighty Cable, manufacturing cables with a cost that varies with distance from the cable's start. The cost function is given as 6√x dollars per meter. Alan calculates the profit from selling a 25-meter cable, which is the revenue (120 * 25) minus the integrated cost from 0 to 25 meters. He then derives a general expression for profit as a function of cable length (K) and finds the maximum profit point by setting the derivative equal to zero, solving for K = 400 meters. However, Alan acknowledges a mistake in calculating the actual maximum profit. The video concludes with a reminder to engage with the content and offers additional help on platforms like Twitch and Discord.
Takeaways
- 📈 The problem involves calculating profit for a company, Mighty Mite, that manufactures cables with a cost that varies with distance from the beginning of the cable.
- 💲 The cost to produce a portion of the cable X meters from the beginning is given by 6√X dollars per meter.
- 🧮 Profit is calculated as revenue minus cost, where revenue is the selling price times the length of the cable sold.
- 🔢 The selling price per meter is constant at $120, but the cost per meter increases with the square root of the distance from the start of the cable.
- ⌛ To find the profit for a 25-meter cable, an integral from 0 to 25 meters of the cost function is required.
- 📊 The integral represents the cumulative cost to produce the cable, which is then subtracted from the total revenue to find profit.
- 🤔 The maximum profit is found by differentiating the profit function with respect to the length of the cable (K) and setting it to zero.
- 🔧 Using the fundamental theorem of calculus, the derivative of the cost function is 6√K, which is set equal to the selling price per meter ($120) to find the maximum profit point.
- 🏁 The maximum profit occurs when the selling price per meter equals the derivative of the cost function, which is at K = 400 meters.
- 🧐 The actual maximum profit is calculated by plugging K = 400 back into the original profit function and evaluating the integral from 0 to 400.
- 📝 The presenter made a mistake in calculating the actual profit and emphasized the importance of checking the calculations against the problem's requirements.
- 📚 The video provides a walkthrough of solving an AP calculus problem, emphasizing the process and the importance of understanding calculus concepts in a practical context.
Q & A
What is the context of the problem discussed in the video?
-The video discusses a problem from the 2009 AP Calculus exam involving a company, Mighty Cable, that manufactures cables. The cost to produce a portion of the cable varies with its distance from the beginning of the cable, and the company's profit is the difference between the revenue from selling the cable and the cost of producing it.
What is the cost function for the cable as described in the video?
-The cost function for the cable is given as 6√x dollars per meter, where x represents the distance in meters from the beginning of the cable.
How is the profit of the company calculated in this context?
-The profit is calculated as the revenue from selling the cable minus the cost of producing it. The revenue is the selling price per meter multiplied by the length of the cable sold, and the cost is the integral of the cost function from 0 to the length of the cable sold.
What is the selling price per meter of the cable?
-The selling price per meter of the cable is $120.
How does the video calculate the profit for a 25-meter cable?
-The profit for a 25-meter cable is calculated by multiplying the selling price per meter ($120) by the length (25 meters) to get the revenue, and then subtracting the integral of the cost function from 0 to 25 meters.
What is the integral expression used to calculate the cost of producing a cable of length K meters?
-The integral expression used to calculate the cost is ∫ from 0 to K of 6√x dx, which represents the accumulated cost per meter over the length of the cable from 0 to K meters.
How does the video attempt to find the maximum profit?
-The video attempts to find the maximum profit by taking the derivative of the profit function with respect to K (the length of the cable) and setting it equal to zero, then solving for K to find the length that maximizes profit.
What is the maximum length of the cable that can be sold for maximum profit according to the video?
-The maximum length of the cable that can be sold for maximum profit is 400 meters, as determined by setting the selling price equal to the derivative of the cost function and solving for K.
What mistake does the presenter make in the video when calculating the actual maximum profit?
-The presenter mistakenly calculates the integral to find the cost but forgets to actually compute the final profit by substituting the value of K (400 meters) back into the profit function.
What is the correct way to calculate the actual maximum profit?
-The correct way to calculate the actual maximum profit is to substitute the value of K (400 meters) back into the profit function and evaluate the integral of the cost function from 0 to 400 meters, then subtract this cost from the revenue (120 * 400).
What is the added cost to go from a 25-meter cable to a 30-meter cable as discussed in the video?
-The added cost to go from a 25-meter cable to a 30-meter cable is calculated by evaluating the integral of the cost function from 25 to 30 meters and adding it to the cost of producing a 25-meter cable.
What is the presenter's recommendation for viewers to get more help with similar problems?
-The presenter offers free homework help on platforms like Twitch and Discord and encourages viewers to comment, like, or subscribe for more content.
Outlines
📈 Calculating Profit from Cable Production
In this segment, Alan discusses a problem from the 2009 AP Calculus exam concerning a cable company's profit from manufacturing and selling cables. The company incurs a variable cost to produce a cable, which is described by a function of the distance from the cable's start. Alan explains how to calculate the profit from selling a 25-meter cable, which involves integrating the cost function over the length of the cable. He also derives an expression for the profit as a function of the cable length and uses calculus to find the maximum profit the company can earn from selling one cable. The key insight is that profit is maximized when the selling price per meter equals the derivative of the cost function, leading to an optimal cable length of 400 meters.
🧮 Profit Analysis and Maximum Profit Calculation
Alan continues the discussion on the cable company's profit, focusing on the difference in cost between producing a 30-meter and a 25-meter cable. He acknowledges a mistake in not calculating the actual maximum profit, which is a critical step in solving the problem. He then attempts to correct this by calculating the profit for a 400-meter cable, integrating the cost function over the length and subtracting it from the revenue. The explanation includes a minor error in the integral calculation, which Alan catches and corrects. The video ends with Alan encouraging viewers to engage with the content and offering additional help through his online platforms.
Mindmap
Keywords
💡AP Calculus Exam
💡Free Response Question
💡Profit
💡Cost Function
💡Integral
💡Revenue
💡Derivative
💡Fundamental Theorem of Calculus
💡Maximum Profit
💡Root Function
💡Problem Solving
Highlights
Alan introduces the 3rd free response question from the 2009 AP Calculus exam.
Mighty Cable Company manufactures cables at a cost varying with distance from the beginning of the cable.
Cost to produce a portion of cable X meters from the beginning is given as 6√X dollars per meter.
Profit is defined as the difference between revenue from selling the cable and the cost to produce it.
Revenue from selling a 25-meter cable is calculated as 120 * 25.
To find the cost, Alan suggests integrating the cost rate from 0 to 25 meters.
Alan uses a calculator to integrate the cost function from 0 to 25 meters.
Profit for a 25-meter cable sale is calculated to be $2,500.
For part B, Alan writes an expression for the profit on the sale of a K-meter long cable.
The profit function is derived by subtracting the cost integral from the revenue.
To find the maximum profit, Alan takes the derivative of the profit function with respect to K.
Setting the derivative equal to zero gives the point where profit is maximized.
The maximum profit point is found where the selling rate equals the cost rate, which is at K = 400 meters.
Alan makes a mistake in calculating the actual maximum profit and corrects the process.
The integral from 0 to 400 of 6√X is calculated to find the cost at the maximum profit point.
Alan corrects his previous error and recalculates the profit using the correct integral.
The video concludes with Alan offering free homework help on Twitch and Discord.
Alan encourages viewers to comment, like, or subscribe for more content.
Transcripts
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