Question 204444
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The perimeter of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P=2l + 2w]


which can be restated as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l = \frac{P - 2w}{2}]


The area of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A = lw]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w) = \left(\frac{P - 2w}{2}\right)w = \frac{Pw - 2w^2}{2}]


This function has an extremum where the first derivative is equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dw} = \frac{P}{2} - 2w]


Setting the first derivative equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{P}{2} - 2w = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{P}{2} = 2w]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w = \frac{P}{4}]


This extremum is a maximum if the value of the 2nd derivative is negative.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{dw^2} = -2]


Which is negative for all values of the independent variable, hence the extremum is a maximum.


In order for the width to be one fourth of the perimeter, the length must also be one-fourth of the perimeter.  That is to say, the maximum area for a rectangle with a given perimeter is achieved when the rectangle is a square with side measure of one-fourth of the perimeter.


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Let *[tex \Large p] represent the price of one fare.  Then the revenue function is the number of passengers times the price of a fare.  The price of one fare minus 36, quantity divided by 2 is the number of $2 increases to the price. So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ R(p) = \left(300 - 10\left(\frac{p-36}{2}\right)\right)p]


Which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 480p - 5p^2]


First derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dR}{dp}=480 - 10p]


Set equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 480 = 10p \ \ \Rightarrow\ \ p = 48]


Second derivative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2R}{dp^2}=-10]


Hence extreme point is a maximum.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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