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In the cozy riverside village of Psychotropic Town, an outbreak of the Angry Duck Eating Soap virus necessitates quarantine. To protect the neighboring villages, a fence needs to be built around town. Using mathematical equations and derivations, determine the optimal fence area with the least material to satisfy the money-grubbing mayor’s request. Explore the Happy F-Bomb Factory scenario to calculate production levels for maximizing profit with cost and demand functions. Learn how derivatives aid in optimizing decisions and solving complex problems.
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A cozy riverside village, Psychotropic Town, is about to be quarantined because of a deadly outbreak of an unnamed virus. The virus is Angry Duck Eating Soap (ADES). In order to protect the neighboring village, you will need to construct a fence around this town. Of course, because the mayor of Psychotropic Town is a money grubbing cretin, he wants you to construct it with 1000 feet of fence and create the largest area possible. Since a river borders one side of the town, you won't need to fence that side off. People with ADES can't swim... duh... Zombie Cancer
x = length y = width A = x*y (Primary) 2x + y = 1000 (Secondary) Variable & Equation Declaration
2x + y = 1000 y = -2x+1000 Solving for width(y)
A = (x)(1000-2x) A = 1000x – 2x22 Creating a New Area Equation
A = 1000x – 2x22 A' = 1000 – 4x 0 = 1000 – 4x -1000 = -4x X = 250 0 Length is 250 feet. Differentiating the Area Equation
1000 = 2x + y 1000 = 2(250) + y 1000 = 500 + y 500 = y Width is 500 feet. Solving for width
Second Derivative Test A'' = -4 (it is concave down, meaning it's a relative maximum) Proving ourselves
Happy F-Bomb factory is making a new product, the H-Bomb. In order for the shipping to commence, an appropriate and environment-friendly cost plan has to be mapped out. Determine the production level that will maximize the profit for the company with cost and demand functions: C(x)=84+1.26x-0.01x2+0.00007x3 and p(x)=3.5-0.01x and What is the maximum profit? Happy F-Bomb Factory
x = units P = profit (R – C) p(x) = 3.5 - .01x C = C(x) = 84+1.26x-.01x2-.00007x3 R = p(x)[x] = 3.5x-.01x2 Variable and Equation Declaration
P = R – C P = 3.5x – 0.01x2 – 84 – 1.26x + 0.01x2 – 0.00007x3 = -0.00007x3 + 2.24x - 84 Solving for Profit
P' = -0.00021x2 + 2.24 x2 = 10666.666 x = 103.28 Solve for x!
P(x) = -0.00007x3 + 2.24x -84 P(103.28) = 70.23 Insert x into equation
P''(x) = -.00042x P''(103.28) = -0.0433776 Negative = Maximum = Concave Up = Yeah Proof of PURCHASE !$#@
Using derivatives allow you to find great deals on products, and how much (or little) of stuff you have to do for the most (or little) of something else! It also allows Mrs. Lustgraaf to derive the maximum possible grade for this presentation!! Conclusion