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optimization.html
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<h1>Optimization</h1>
<p>This section discusses a basic application of calculus to answer questions which relate to the largest or smallest a function can be given some constraints.</p>
<p>For example,</p>
<blockquote>
<p>Of all rectangles with perimeter 20, which has of the largest area?</p>
</blockquote>
<p>The main tool is the extreme value theorem of Bolzano and Fermat's theorem about critical points: If the function <span class="math">$f(x)$</span> is continuous on <span class="math">$[a,b]$</span> and differentiable on <span class="math">$(a,b)$</span>, then the extrema exist and must occur at either an end point or a critical point.</p>
<p>Though not all of our problems lend themselves to a description of a continuous function on a closed interval, if they do, we have an algorithmic prescription to find the absolute extrema of a function:</p>
<ol>
<li><p>Find the critical points. For this we will can use a root-finding function like <code>find_zero</code>.</p>
</li>
<li><p>Evaluate the function values at the critical points and at the end points.</p>
</li>
<li><p>Identify the largest and smallest values.</p>
</li>
</ol>
<p>With the computer we can take some shortcuts, as we will be able to graph our function to see where the extreme values will be.</p>
<h2>Fixed perimeter and area</h2>
<p>The simplest way to investigate the maximum or minimum value of a function over a closed interval is to just graph it and look.</p>
<p>We begin with the question of which rectangles of perimeter 20 have the largest area? The figure shows a few different rectangles with this perimeter and their respective areas.</p>
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<figure>
<img 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<figcaption><div class="markdown"><p>Some possible rectangles that satisfy the constraint on the perimeter and their area.</p>
</div></figcaption>
</figure>
</div>
<p>The basic mathematical approach is to find a function of a single variable to maximize or minimize. In this case we have two variables describing a rectangle: a base <span class="math">$b$</span> and height <span class="math">$h$</span>. Our formulas are the area of a rectangle:</p>
<p class="math">\[
~
A = bh,
~
\]</p>
<p>and the formula for the perimeter of a rectangle:</p>
<p class="math">\[
~
P = 2b + 2h = 20.
~
\]</p>
<p>From this last one, we see that <span class="math">$b$</span> can be no bigger than 10 and no smaller than 0 from the restriction put in place through the perimeter. Solving for <span class="math">$h$</span> in terms of <span class="math">$b$</span> then yields this restatement of the problem:</p>
<p>Maximize <span class="math">$A(b) = b \cdot (10 - b)$</span> over the interval <span class="math">$[0,10]$</span>.</p>
<p>This is exactly the form needed to apply our theorem about the existence of extrema (a continuous function on a closed interval). Rather than solve analytically by taking a derivative, we simply graph to find the value:</p>
<pre class='hljl'>
<span class='hljl-k'>using</span><span class='hljl-t'> </span><span class='hljl-n'>CalculusWithJulia</span><span class='hljl-t'> </span><span class='hljl-cs'># loads `Plots`, `ForwardDiff`, `Roots`, `SymPy`</span><span class='hljl-t'>
</span><span class='hljl-nf'>A</span><span class='hljl-p'>(</span><span class='hljl-n'>b</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>=</span><span class='hljl-t'> </span><span class='hljl-n'>b</span><span class='hljl-t'> </span><span class='hljl-oB'>*</span><span class='hljl-t'> </span><span class='hljl-p'>(</span><span class='hljl-ni'>10</span><span class='hljl-t'> </span><span class='hljl-oB'>-</span><span class='hljl-t'> </span><span class='hljl-n'>b</span><span class='hljl-p'>)</span><span class='hljl-t'>
</span><span class='hljl-nf'>plot</span><span class='hljl-p'>(</span><span class='hljl-n'>A</span><span class='hljl-p'>,</span><span class='hljl-t'> </span><span class='hljl-ni'>0</span><span class='hljl-p'>,</span><span class='hljl-t'> </span><span class='hljl-ni'>10</span><span class='hljl-p'>)</span>
</pre>
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<p>You should see the maximum occurs at <span class="math">$b=5$</span> by symmetry, so <span class="math">$h=5$</span> as well, and the maximum area is then <span class="math">$25$</span>. This gives the satisfying answer that among all rectangles of fixed perimeter, that with the largest area is a square. As well, this indicates a common result: there is often some underlying symmetry in the answer.</p>
<h3>Exploiting polymorphism</h3>
<p>Before moving on, let's see a slightly different way to do this problem with <code>Julia</code>, where we trade off some algebra for a bit of abstraction. This was discussed in the section on <a href="../precalc/functions.html">functions</a>. Let's first write area as a function of both base and height:</p>
<pre class='hljl'>
<span class='hljl-nf'>A</span><span class='hljl-p'>(</span><span class='hljl-n'>b</span><span class='hljl-p'>,</span><span class='hljl-t'> </span><span class='hljl-n'>h</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>=</span><span class='hljl-t'> </span><span class='hljl-n'>b</span><span class='hljl-oB'>*</span><span class='hljl-n'>h</span>
</pre>
<pre class="output">
A (generic function with 2 methods)
</pre>
<p>Here we write area, quite naturally, as a function of two variables.</p>
<p>Then from the constraint given by the perimeter being a fixed value we can solve for <code>h</code> in terms of <code>b</code>. We write this as a function:</p>
<pre class='hljl'>
<span class='hljl-nf'>h</span><span class='hljl-p'>(</span><span class='hljl-n'>b</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>=</span><span class='hljl-t'> </span><span class='hljl-p'>(</span><span class='hljl-ni'>20</span><span class='hljl-t'> </span><span class='hljl-oB'>-</span><span class='hljl-t'> </span><span class='hljl-ni'>2</span><span class='hljl-n'>b</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>/</span><span class='hljl-t'> </span><span class='hljl-ni'>2</span>
</pre>
<pre class="output">
h (generic function with 1 method)
</pre>
<p>Then to get <code>A(b)</code> we simply need to substitute <code>h(b)</code> into our formula for the area, <code>A</code>. However, instead of doing the substitution ourselves using algebra we let <code>Julia</code> do it through composition of functions:</p>
<pre class='hljl'>
<span class='hljl-nf'>A</span><span class='hljl-p'>(</span><span class='hljl-n'>b</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>=</span><span class='hljl-t'> </span><span class='hljl-nf'>A</span><span class='hljl-p'>(</span><span class='hljl-n'>b</span><span class='hljl-p'>,</span><span class='hljl-t'> </span><span class='hljl-nf'>h</span><span class='hljl-p'>(</span><span class='hljl-n'>b</span><span class='hljl-p'>))</span>
</pre>
<pre class="output">
A (generic function with 2 methods)
</pre>
<p>From this we can solve graphically as before, or numerically. We search for zeros of the derivative:</p>
<pre class='hljl'>
<span class='hljl-nf'>find_zeros</span><span class='hljl-p'>(</span><span class='hljl-n'>A</span><span class='hljl-oB'>'</span><span class='hljl-p'>,</span><span class='hljl-t'> </span><span class='hljl-ni'>0</span><span class='hljl-p'>,</span><span class='hljl-t'> </span><span class='hljl-ni'>10</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-cs'># find_zeros in `Roots`,</span>
</pre>
<pre class="output">
1-element Array{Float64,1}:
5.0
</pre>
<div class="alert alert-success" role="alert">
<div class="markdown"><p>Look at the last definition of <code>A</code>. The function <code>A</code> appears on both sides, though on the left side with one argument and on the right with two. These are two "methods" of a <em>generic</em> function, <code>A</code>. <code>Julia</code> allows multiple definitions for the same name as long as the arguments (their number and type) can disambiguate which to use. In this instance, when one argument is passed in then the last defintion is used (<code>A(b,h(b))</code>), whereas if two are passed in, then the method that multiplies both arguments is used. The advantage of multiple dispatch is illustrated: the same concept–-area–-has one function name, though there may be different ways to compute the area, so there is more than one implementation.</p>
</div>
</div>
<h3>Norman Windows</h3>
<p>Here is a similar, though more complicated, example where the analytic approach can be a bit more tedious, but the graphical one mostly satisfying, though we do use a numerical algorithm to find an exact final answer.</p>
<p>Let a "<a href="https://en.wikipedia.org/wiki/Norman_architecture">Norman</a>" window consist of a rectangular window of top length <span class="math">$x$</span> and side length <span class="math">$y$</span> and a half circle on top. The goal is to maximize the area for a fixed value of the perimeter. Again, assume this perimeter is 20 units.</p>
<p>This figure shows two such windows, one with base length given by <span class="math">$x=4$</span>, the other with base length given by <span class="math">$x=3$</span>. The one with base length <span class="math">$4$</span> seems to have much bigger area, what value of <span class="math">$x$</span> will lead to the largest area?</p>
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<p>For this problem, we have two equations.</p>
<p>The area is the area of the rectangle plus the area of the half circle (<span class="math">$\pi r^2/2$</span> with <span class="math">$r=x/2$</span>).</p>
<p class="math">\[
~
A = xy + \pi(x/2)^2/2
~
\]</p>
<p>In <code>Julia</code> this is</p>
<pre class='hljl'>
<span class='hljl-nf'>A</span><span class='hljl-p'>(</span><span class='hljl-n'>x</span><span class='hljl-p'>,</span><span class='hljl-t'> </span><span class='hljl-n'>y</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>=</span><span class='hljl-t'> </span><span class='hljl-n'>x</span><span class='hljl-oB'>*</span><span class='hljl-n'>y</span><span class='hljl-t'> </span><span class='hljl-oB'>+</span><span class='hljl-t'> </span><span class='hljl-n'>pi</span><span class='hljl-oB'>*</span><span class='hljl-p'>(</span><span class='hljl-n'>x</span><span class='hljl-oB'>/</span><span class='hljl-ni'>2</span><span class='hljl-p'>)</span><span class='hljl-oB'>^</span><span class='hljl-ni'>2</span><span class='hljl-t'> </span><span class='hljl-oB'>/</span><span class='hljl-t'> </span><span class='hljl-ni'>2</span>
</pre>
<pre class="output">
A (generic function with 2 methods)
</pre>
<p>The perimeter consists of 3 sides of the rectangle and the perimeter of half a circle (<span class="math">$\pi r$</span>, with <span class="math">$r=x/2$</span>):</p>
<p class="math">\[
~
P = 2y + x + \pi(x/2) = 20
~
\]</p>
<p>We solve for <span class="math">$y$</span> in the first with <span class="math">$y = (20 - x - \pi(x/2))/2$</span> so that in <code>julia</code> we have:</p>
<pre class='hljl'>
<span class='hljl-nf'>y</span><span class='hljl-p'>(</span><span class='hljl-n'>x</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>=</span><span class='hljl-t'> </span><span class='hljl-p'>(</span><span class='hljl-ni'>20</span><span class='hljl-t'> </span><span class='hljl-oB'>-</span><span class='hljl-t'> </span><span class='hljl-n'>x</span><span class='hljl-t'> </span><span class='hljl-oB'>-</span><span class='hljl-t'> </span><span class='hljl-n'>pi</span><span class='hljl-t'> </span><span class='hljl-oB'>*</span><span class='hljl-t'> </span><span class='hljl-n'>x</span><span class='hljl-oB'>/</span><span class='hljl-ni'>2</span><span class='hljl-p'>)</span><span class='hljl-t'> </span><span class='hljl-oB'>/</span><span class='hljl-t'> </span><span class='hljl-ni'>2</span>
</pre>