Consider an isosceles triangle with sides 5,5, and 6. Find the dimensions of the rectangle of largest area that can be inscribed in the triangle so that one side is along the base.
sketch the triangle with vertex on the y-axis and base on the x-axis.
two 3-4-5 triangles are formed in quads I and II
using the first quadrant triangle, the line formed by side of length 5 has the equation $\displaystyle y = -\frac{4}{3}x+4$
consider a rectangle with upper right corner on the this line
area of a rectangle in quad I , $\displaystyle A = x\left(-\frac{4}{3}x+4\right)$
maximize this one, and using symmetry, double the result to get the overall max area.
let ABC be an isosceles triangle with AB=AC=5 and BC=6.
let angle ABC=a
let length of sides of rectangle be x and y
A=area of rectangle
A=xy
tan(a)=2y/(6-x)
y=(6-x)tan(a)/2
putting this value we get
A=[tan(a)/2]x(6-x)
A=[tan(a)/2](6x-x^2)
differentiating we get
A'=[tan(a)/2](6-2x)
equating it to 0 we have
[tan(a)/2](6-2x)=0
or x=3
hence for maximum value x=3
now y may be found from equatin 2y=tan(a)(6-x)
or 2y=3tan(a)
value of tan(a)=4/3(draw perpendicular from A on BC and use pythagoras theorem)
so y=2
hence x=3 and y=2