Question 403095
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Let *[tex \Large y] represent the measure of the vertical dimension of the rectangular portion of the window.  Let *[tex \Large x] represent the horizontal dimension of the window.  *[tex \Large x] is also the diameter of the semi-circular part of the window.


The perimeter of the window is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(x,y)\ =\ 2y\ +\ x\ +\ \frac{\pi x}{2}]


But we are given that the perimeter is 20 feet, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2y\ +\ x\ +\ \frac{\pi x}{2}\ =\ 20]


Solve for *[tex \Large y]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 10\ -\ \frac{(2\ +\ \pi)x}{4} ]


The area of the window is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x,y)\ =\ xy\ +\ \frac{\pi{\frac{x}{2}}^2}{2}]


Substitute to create a function for area in terms of *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ 10x\ -\ \frac{(2\ +\ \pi)x^2}{4}\ +\ \frac{\pi{\frac{x}{2}}^2}{2}]


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ 10x\ -\ \frac{(4\ +\ \pi)}{8}x^2]


Two ways to go from here:


<u><b>Algebra Method</b></u>


This is a quadratic with a negative lead coefficient, hence the graph is a parabola that opens downward.  That means the vertex is a maximum.  Recall that the vertex of *[tex \Large \rho(x)\ =\ ax^2\ +\ bx\ +\ c] is at *[tex \Large x_v = \frac{-b}{2a}].


For this problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_{max} = \frac{-10}{-\frac{(4\ +\ \pi)}{4}}\ =\ \frac{40}{4\ +\ \pi}]


and a little arithmetic gets us to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_{max}\ =\ \frac{10(2\ +\ \pi)}{4\ +\ \pi}]


And the radius of the semi-circle is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x_{max}}{2} = \frac{20}{4\ +\ \pi}]


<u><b>Calculus Method</b></u>


This is a 2nd degree polynomial equation, hence continuous and twice differentiable over the real numbers.  Therefore there will be a local extremum at any point where the first derivative is equal to zero and that point will be a maximum if the 2nd derivative is negative at that point.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dx}\ =\ 10\ -\ \frac{(4\ +\ \pi)}{4}x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\ -\ \frac{(4\ +\ \pi)}{4}x\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{40}{4\ +\ \pi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{dx^2}\ =\ -\ \frac{(4\ +\ \pi)}{4}\ <\ 0]


Hence *[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{40}{4\ +\ \pi}]


is the *[tex \Large x]-coordinate of the maximum point.


The *[tex \Large y] dimension and the semi-circle radius are calculated as above.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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