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Optimization for the Awesome Calculus Student
Okay, so you’re a Calculus 1 student
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You know the
trig functions and know the unit circle like the days of the week
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Now that you know how to derive using the product, chain,
quotient, or power rules and have an understanding of the first and second derivative and what
they tell you, it is time for some practical application
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It is
a fundamental, practical use of derivatives that sees daily use throughout the Calculus world
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Optimization is using derivatives in order to find the largest or smallest value of a function given
certain constraints
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The
basic structure of an optimization problem is something like this: two functions are employed
using given information, substitution and derivatives are employed, and a final solution fitting
the given constraints is found and verified to be a functional maximum or minimum
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There is a field that is of needing to be enclosed
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Also, there is a building present on one side of the field so
no fence needs to go up there
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Here we are given the task of making a fence
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Draw a Picture
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And since Calculus students
obviously have wonderful imaginations, we can focus on the task of what is in this field
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Building (No Fence)

Y

Y

x
Figure 1: What My Field Would Look Like
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Create Equations with Given Information
We know that the area of this given fence has to be the length times the width
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Thus, we have two equations!

A = xy & 500 = x + 2y
We are going to be utilizing the area function to find the maximum area, but we can’t derive the
given with two different variables
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Substitution Using the Constraint Function

We know that if we isolate x in the constraint equation, we can substitute that value for x in the
other equation
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500 = x + 2y
x = 500 – 2y
Plugging this into area we get
A = (500-2y)y or A= 500y-2
Now that we have this, we can…

4
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Thus, we need to put down a domain that is applicable for the given constraints
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So we derive and
set equal to zero and if any value is not within this domain, we dispose of it
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Set Derivative Equal to Zero and Find Critical Point(s)
Solving for the A’ by setting it equal to zero gets y = 125
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Basically, we just utilized the first derivative to find a critical point in the
function of A = -2

+ 500y
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For

more complicated problems, the second derivative test can be employed to ensure that the value
obtained is the maximum or minimum
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And that brings us to the final step
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Use the Solved Value to Find the Other Dimension(s)
y= 125
500 = x + 2(125)
x = 250
So, the dimensions to give the largest area are 125 feet by 250 feet
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Always look back at the question to verify what was being asked!
In conclusion, to simplify the given work, the basic steps for Optimization are…

1
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Create Equations with Given Information
3
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Determine the Domain and Derive
5
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Use the Solved Value to Find the Other Dimension(s)
Remember to take your time on problems, think over the given information, and make the
problem your own! Personalize the picture and have fun; it’s what calculus is all about!



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