I encountered this problem while performing precalculus homework. I need to maximize for the area of the rectangle in the following figure with a perimiter of 400 units:
Obviously, the first steps are to form equations for the perimeter and area in terms of one variable:
$$P = 2x + \pi y = 400 \\ y = \frac {400-2x} {\pi } $$ Substituting for $y$ in the area for the rectangle, $xy$ $$ A = x \left( \frac {400-2x} {\pi } \right) $$
Great! Now I have a quadratic equation inverted over the x-axis in the form of ($ax^2 + bx)$, so the vertex is the maximum.
$\frac {-2}{\pi} x^2 + \frac {400}{\pi} x$
Plugging these values into the formula for vertex ($\frac {-b} {2a}$), derived from applying the midpoint formula to the quadratic equation yields $$ \frac {\frac {-400}{\pi}} {\frac {-4}{\pi}} = \frac {-400}{\pi} \times \frac{\pi}{-4} = \frac {-400 \pi} {-4 \pi} = 10$$
However, the online homework program has marked $x=10$ as incorrect. I know that my perimeter and area equations are correct, and I cannot seem to find any arithmetic errors. Perhaps I am using an incorrect method altogether and should apply something else? I'm not sure specifically what to ask because I've exhausted the methods I know.
Just a careless mistake.
$$\frac{400}{4}=100$$
You have a missing $0$.