What should be the length of the third side of an isosceles triangle, if the congruent sides is 10 and the area is at maximum?
Solution:
\(A = \frac{bh}{2}\)
\(10^2 = h^2 +(\frac{b}{2})^2\)
\(h = \sqrt{100- (\frac{b}{2})^2}\)
\(A = \frac{(b)\sqrt{100- (\frac{b}{2})^2}}{2}\)
\(A = \frac{\sqrt{100b^2- \frac{b^4}{4}}}{2}\)
\(\frac{2dA}{dx} = \frac{200b-b^3}{2\sqrt{100b^2-\frac{b^4}{4}}}\)
\(0 = 200b-b^3\)
\(b = 14.14 \) units
Thus the length of the third side is:
\(b = 14.14 \) units