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Chapter 3: Applications of Differentiation
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Section 3.8: Optimization
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Example 3.8.1


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Of all rectangles with perimeter 100, find the one with maximal area.



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Solution



Analysis


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Figure 3.8.1(a) shows a labeled rectangle whose area is $f\left(w\,h\right)\=wh$, and whose perimeter is $2wplus;2h$.

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The constraint equation is $g\left(w\,h\right)\equiv 2wplus;2h100equals;0$.

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Solve the constraint equation for, say, $h\=50w$ and write the area of the rectangle as $w\left(50w\right)$.

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Maximize the objective function $F\left(x\right)\=x\left(50x\right)$${}$


>

p1:=plottools[rectangle]([1,4],[5,1],style=line):
p2:=plots:textplot({[3,.7,typeset(w)],[.7,2.5,typeset(h)]},font=[Lucinda,18]):
plots:display(p1,p2,scaling=constrained, axes=none);


Figure 3.8.1(a) Labeled diagram of a rectangle






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Graphical Solution


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Figure 3.8.1(b) is a graph of the objective function $F\left(x\right)\=x\left(50x\right)$. The curve is a parabola, and the vertex of the parabola is the absolute maximum for the function $F$.

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Using the Context Panel for the graph, set Probe Info to Nearest datum, and trace the graph with the crosshair form of the probe.

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Set the crosshair on the vertex. Again in the Context Panel for the graph, in the Probe Info option, select Copy data.

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Paste the contents of the Clipboard into the workspace:

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$\left[\begin{array}{c}25.11518285929648\\ 624.9867329089243\end{array}\right]$



Figure 3.8.1(b) Graph of $F\left(x\right)\=x\left(50x\right)$






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The density of pixels on the computer screen determines how accurately a graph can be "read." From the graphical approximation, it might be conjectured that the optimal dimensions of the rectangle are $25\times 25$, for a maximal area of 625.${}$
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Numeric Solution


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Numeric solution via the Context Panel

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Form a sequence of the objective function and the constraint equation.

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Context Panel: Optimization≻Maximize (local)


$whcomma;2wplus;2hequals;100$$\stackrel{\text{maximize}}{\to}$$\left[{624.999999999999}{\,}\left[{h}{\=}{25.0000000000000}{\,}{w}{\=}{25.0000000000000}\right]\right]$



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The return consists of a list with two objects. The first object is the optimal value of the objective function; the second, a list of the parameter values giving this extreme value.
Numeric solution via the Optimization Assistant

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Form a sequence of the objective function and the constraint equation.

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Context Panel:
Optimization≻Optimization Assistant
(The Optimization Assistant launches with the objective function and constraint equation inserted into the appropriate fields.)

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In the Options section, select Maximize

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Click the Solve button. (See Figure 3.8.1(c).)

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Click the Quit button to exit the Optimization Assistant and have it write the Solution to the underlying worksheet.

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Click
to launch the Optimization Assistant with the data embedded.



Figure 3.8.1(c) Solution by Optimization Assistant






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Algebraic Solution


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Controldrag $F\left(x\right)\=\dots$

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Context Panel: Assign Function


$F\left(x\right)\=x\left(50x\right)$$\stackrel{\text{assign as function}}{\to}$${F}$

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Type $F\left(x\right)$ and press the Enter key.

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Context Panel: Expand≻Expand

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Context Panel: Complete Square≻$x$


$F\left(x\right)$
${x}{}\left({50}{}{x}\right)$
$\stackrel{\text{expand}}{\=}$
${}{{x}}^{{2}}{\+}{50}{}{x}$
$\stackrel{\text{complete square}}{\=}$
${}{\left({x}{}{25}\right)}^{{2}}{\+}{625}$
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Upon completing the square in $F\left(x\right)$, the equation $y\={\left(x25\right)}^{2}\+625$ describes a parabola in vertex form. The coordinates of the vertex can be read from the equation: $\left(25\,625\right)$. Hence, the width of the rectangle is 25, its corresponding height is $\left(5025\right)\=25$, and the maximal area is 625. The rectangle with fixed perimeter and maximal area is a square.${}$${}$
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Analytic Solution


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Write the equation for critical numbers.
Press the Enter key.

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Context Panel: Solve≻Solve


$F\prime \left(x\right)\=0$
${50}{}{2}{}{x}{\=}{0}$
$\stackrel{\text{solve}}{\to}$
$\left\{{x}{\=}{25}\right\}$

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Determine the maximal area.


$F\left(25\right)$ = ${625}$${}$



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If $w\=25$, then the constraint equation $2wplus;2hequals;100$ determines that $h\=25$, so the rectangle whose perimeter is fixed at 100, and whose area is a maximum, is a square.${}$
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The purist might demand that the SecondDerivative test be applied via the calculation $F\u2033\left(25\right)\=2$, but the pragmatist will insist that Figure 3.8.1(b) is sufficient.



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