Really need a Calc expert! A rectangle (positioned askew) inscribed in a right triangle, maximize area?
A rectangle is to be inscribed in a right triangle having sides of length 6(ground leg), 8 (standing leg), and 10 in (hypotenuse). Find the dimensions of the rectangle with the greatest area assuming the rectangle is positioned with one side running directly on the hypotenuse (so like tipped up on one side). Need explanation! But any ideas will help.
This one is hard to explain in print. Let x be the side of the rectangle on the hypotenuse and y be the other side. There is a triangle formed at the right angle of the original which is similar to the original.
So x is to 10 as z is to 8: x/10 = z/8
and z = 4/5 x
The upper part of that leg would be 8 – 4/5 x
The triangle at the top is also similar:
So y is to 6 as (8 – 4/5 x) is to 10: y/6 = (8-4/5 x)/10
and y = 3/5 (8- 4/5 x) = 24/5 – 12/25 x
Now area = xy
Deriv = 0 and solve
I called the side along the hypotenuse x.
Using similar triangles, the other side is 4.8 -.48x
(The smaller 3-4-5 triangle with the original right angle and hypotenuse x, has longer side .8x and short side .6x
area function =x(4.8 -.48x)
Either find the vertex or take first derivative and you get x=5
(the length) and width of rectangle =2.4
The area is 12, which is half the area of the original triangle.
References :
This one is hard to explain in print. Let x be the side of the rectangle on the hypotenuse and y be the other side. There is a triangle formed at the right angle of the original which is similar to the original.
So x is to 10 as z is to 8: x/10 = z/8
and z = 4/5 x
The upper part of that leg would be 8 – 4/5 x
The triangle at the top is also similar:
So y is to 6 as (8 – 4/5 x) is to 10: y/6 = (8-4/5 x)/10
and y = 3/5 (8- 4/5 x) = 24/5 – 12/25 x
Now area = xy
Deriv = 0 and solve
References :