Let the height of the triangle be [tex] h [/tex] and the base be [tex] b [/tex].
Given that [tex] h+b=26 [/tex] and the area of the triangle is [tex] A=\frac{bh}{2} [/tex]. Or
[tex] A(h)=\frac{(26-h)h}{2} [/tex].
When the area is maximum, [tex] \frac{dA}{dh} =0 [/tex].
or,
[tex] \frac{d\frac{(26-h)h}{2} }{dh} =0 [/tex].
[tex] \frac{d(26-h)h }{dh} =0 [/tex]
[tex] 2h-26=0 [/tex]
[tex] h=13 [/tex]
The dimensions of the triangle with maximum area is
[tex] h=13 cm, b=13 cm [/tex]
