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?

1 Answer
Mar 6, 2018

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.