Question 337695
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Typing in all caps is the electronic equivalent of shouting.  It is therefore both rude and annoying.  Inside voice boys and girls, please.


Let *[tex \Large x] represent one of the numbers.  Then the other number is *[tex \Large 96\ -\ x]


Then we want to maximize *[tex \Large f(x)\ =\ 96x\ -\ x^2] where *[tex \Large x\ \in\ \mathbb{N}]


<b><u>Calculus solution:</u></b>


Set the first derivative of the function equal to zero and solve for the *[tex \Large x]-coordinate of the extreme point.  Evaluate the second derivative at the calculated value of *[tex \Large x] to verify that the extreme is a maximum.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}f(x)\ =\ 96\ -\ 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 96\ -\ 2x\ =\ 0\ \ \Rightarrow\ \ x\ =\ 48]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2}{dx^2}f(x)\ =\ -2\ <\ 0\ \forall x\ \in\ \mathbb{R}]


Therefore, the function has a maximum at *[tex \Large (48, 48^2)]


<b><u>Algebra solution:</u></b>


Note that *[tex \Large f(x)\ =\ 96x\ -\ x^2] is a quadratic polynomial in *[tex \Large x] with a negative lead coefficient.  That means the graph is a parabola opening downward -- and the vertex is the maximum of the function.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -1\ \ ], *[tex \LARGE b\ =\ 96\ \ ], and *[tex \LARGE c\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ x_v\ =\ \frac{-b}{2a}\ =\ \frac{-96}{2(-1)}\ =\ 48]


Same answer.


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