Find the area of the rectangle as a function of its height. What is the maximum possible area of all such rectangles?A rectangle is inscribed in a right isosclese triangle with a hypotenuse of 18...
Find the area of the rectangle as a function of its height. What is the maximum possible area of all such rectangles?
A rectangle is inscribed in a right isosclese triangle with a hypotenuse of 18 units.
Place the right triangle with the right angle at the origin and the sides along the positive x and y axes. Since the hypotenuse has length 18 the side lengths are `9sqrt(2)` . (From the Pythagorean theorem: `x^2+x^2=18^2==>2x^2=324==>x^2=162 ==>x=sqrt(162)=9sqrt(2)` )
Assuming that we have a vertex of the rectangle at the origin, let the height be x. Then the width is the y-coordinate of the hypotenuse at x. The hypotenuse is the line `y=-x+9sqrt(2)` , so the y-coordinate is `-x+9sqrt(2)` .
Thus the height of the rectangle is x, and the width is `-x+9sqrt(2)`
The area is `A=x(-x+9sqrt(2))`
The graph of the area function is a parabola opening down. The maximum occurs at the vertex which is found on the axis of symmetry `x=(-b)/(2a)`
In standard form we have `A=-x^2+9sqrt(2)x` so the axis of symmetry, and the x-coordinate of the vertex, is `x=(-9sqrt(2))/(-2)=4.5sqrt(2)` . If `x=4.5sqrt(2)` then `y=4.5sqrt(2)` .
The area is `4.5sqrt(2)(4.5sqrt(2))=40.5` square units.
As it is a right isoceles triangle therefore the side of the triangle is equal to sqrt(18/2) = 3 units
Let h be the hieght of the inscibed rectangle then the base of the rectangle = 3-h, can be clearly understood of a drawn on a piece of paper
Let a be the area of the rectangle then
f(A)=h(3-h), area of triangle as function of hieght
f(A) = 3h-h^2
f'(A) = 3-2h
for maxima or minima, 3-2h=0 or h=1.5 units
f''(A) = -2, which is negative and shows that the above value is for maxima
The area is maximum when hieght of the rectagle is 1.5 units and the maximum possible area = 1.5(3-1.5) = 2.25 unit^2
The maximum possible area of all such rectangles is 2.25 unit^2