a and b are real numbers and satisfy ab^2 = 27/5 and a^2 b = 1080. Compute a+5b.
ab^2 = 27/5....................(1)
a^2 b = 1080..................(2)
From (2), we have: a ==sqrt [1080 / b]
Sub this into (1) above and solve for b:
Solve for b:
(6 sqrt(30))/(1/b)^(3/2) = 27/5
Take the reciprocal of both sides:
(1/b)^(3/2)/(6 sqrt(30)) = 5/27
Multiply both sides by 6 sqrt(30):
(1/b)^(3/2) = (10 sqrt(10/3))/3
Raise both sides to the power of two:
1/b^3 = 1000/27
Take the reciprocal of both sides:
b^3 = 27/1000
Take the cube root of both sides:
b ==3 / 10 and a==60
a + 5b ==60 + 15/10 ==123/2 ==61.50
