Optimization?

Question

you are planning to close off a corner of the first quadrant with a line of constant length, running from (a,0) and (0,b). Show that the area of the triangle enclosed by the segment is largest when a=b

Answer 1

The area of the triangle is A = .5*a*b. If the line of closure is L, then a and b are related: a^2+b^2=L^2. Make a substitution for b:

a^2 + 4A^2/a^2 = L^2; solve for A: A = (a/2)√[L^2 - a^2] = (1/2)√[L^2*a^2 - a^4]

take the derivative of A wrt a, .5/√[L^2*a^2 - a^4] * (2*a*L^2 - 3*a^3) and set it to zero:

(2*a*L^2 - 4*a^3) = 0

L^2 - 2*a^2 = 0 a = L/√2

Then from a^2+b^2=L^2 (L^2)/2 +b^2 = L^2

this gives b = L/√2 = a

You know the derivative = 0 gives a maximum and not a minimum because the minimum area is when a=0 or b=0 (the area is then 0).