Question

Find the dimensions of the rectangle of maximum area that can be inscribed in a right triangle with base 6 units and height 4 units?

Answer 1

Rectangle will be `3` units long and have height of `2` units.

Explanation:

Consider the area of the enclosing triangle, since it is a right triangle it's area is given by `A=[BH]/2`= `12units^2` in this case.
Let the area of the rectangle =`xy`............`[1]`

Consider the two small triangles left over by the intrusion of the rectangle.

The one lying on the `x` axis will have dimensions `[6-x]y/2`
The one on the `y` axis will have dimensions `[4-y]x/2`

So the total area of the triangle =`12=xy+[6-x]y/2+[4-y]x/2`

Multiplying both sides by `2` and expanding the brackets......

`24=2xy+6y-xy+4x-xy`. i.e ....`24 =6y+4x` and so

`y=[24-4x]/6` and substituting this value for `y` in.......`[1]`

`xy` = `[x][24-4x]/6`..=`24x/6-4x^2/6` = `[4x-2x^2/3]`.

We can now differentiate this with respect to the area A. [since we have the variables in terms of of `x`]

`dA/dx=4-4x/3`=0 for max or min, therefore `x=3`. To see if this value of `x` represents a maximum turning point , and thus maximises the Area in terms of `x` we need the second derivative test.

`d^2A/[dx^2`=`-4/3` which is negative whatever the value of x and so represents a maximum turning point.

Substituting this value of `x`=`3` in `y = [24-4x]/6`, `y =2`

Sorry I am unable to include a sketch, getting on a bit now and technology has left me behind somewhat.