A lot is in the the form of right triangle with perpendicular sides of 60 ft and 80 ft. What are the dimensions of the rectangle with maximum area that can be placed in the triangle.
The lot is in the shape of a triangle with perpendicular sides 60 ft and 80 ft. Let the length of the rectangle with the maximum area be l and it lies along the side with length 80 and the width of the rectangle be w.From the length of the perpendicular sides of the triangle the angle made by the hypotenuse with the horizontal is `A = tan^-1(60/80)`
The ratio of the sides of the triangle is w/(80 - l) and this is equal to tan A or `w/(80 - l) = 60/80`
=> 80w = 4800 - 60l
=> w = 60 - 3l/4
The area of the rectangle is A = w*l = (60 - 3l/4)*l
To maximize A, solve A' = 0
=> 60 - (3*2*l)/4 = 0
=> l = 60*2/3 = 40
w = 30
The width and length of the required rectangle is 30 ft and 40 ft.