u are planning to close off a corner of the 1st quadrant w/ a line segment 20 units long running from (a,0) to

Question

(0,b). How do u show that the triangle enclosed by the segment is lagest wen a=b?

Answer 1

in 1st quadrant, the equation of line in intercepts (x, y axes) passing from (a,0) and (0,b) is
x/a + y/b =1 or y = b - bx/a
area encosed by strt. line within triangle with axes

A = integral y.dx (limit x = to a)
= bx - (b/2a) x^2
= b[a-0] - (b/2a) [a^2 -0]
= ab - ab/2 = ab/2

A= ab/2 (even one can directly write area of triangle)
given a^2 + b^2 =(20)^2 =400 ----(1)

A = (1/2) a * (400 - a^2)^1/2----(2)

dA/da =1/2[(400 - a^2)^1/2 - a (1/2) 2a /(400-a^2)^1/2]
dA/da =[200 - a^2] / (400-a^2)^1/2 ....(3)

for for A largest dA/da = 0 and d^2 A/da^2 = -ve
[200 - a^2] = 0
a^2=200
b^2=200 from (1)
so a=b

d^2 A/da^2 =
[-2a*(400-a^2)^1/2 - (-a) (200-a^2) (400-a^2)^-1/2] /(400-a^2)
= [-2a* (400-a^2) + a(200-a^2)] / (400-a^2)^3/2
= a [-800+a^2 + 200-a^2)] / (400-a^2)^3/2

d^2 A/da^2 = - 600 a / (400-a^2)^3/2 at a = 10 (2)^1/2
d^2 A/da^2 = - 6000 *(2)^1/2 / (200)^3/2
= -ve
proved