A corner lot has the form of a right triangle; it has perpendicular sides measuring 60 m and 80 m,
respectively. It is required to construct the largest rectangular building with sides parallel to the street.
What is the largest area of the building?
+130116 See the following image :
Let x be the width of such a building
The segment BC has a slope of (-80) / (60) = -8/6 = -4/3
And the equation of a line theorugh this segmnt has the equation y = (-4/3) ( x - 60)
Note that at any point x, the length of the building will be equal to this "y"
So.....the area of such a building is x * y = x [ ( -4/3) (x -60) ]
So
Area = (-4/3) (x) ( x -60) = (-4/3)x^2 + 80x
This is a parabola that turns downward
The "x" that will maximize the area = the x coordinate of the vertex = -80 / (2 *-4/3) = 80 / (8/3) = 240 / 8 = 30 m
And the max area is (-4/3)(30)^2 + 80 (30) = -1200 + 2400 = 1200 m^2
{The building is denoted by ADFE}
