+130511 I assume that you have this x√(9-x) = x*(9-x)^(1/2)
Taking the first derivative, we have
(9-x)^(1/2) - (1/2)x(9-x)^(-1/2) Setting to 0 and factoring, we have
(9-x)^(-1/2) [ (9-x) - (1/2)x] = 0
[9 - (3/2)x] = 0
9 = (3/2)x
x = 6
We can show that this is a max by taking the second derivative =
[3(x-12)] / [4(9-x)^(3/2)]
And plugging in the critical point x = 6, we see that the numerator < 0 and the denominator > 0 = negative, so this tells us that (6, 6√3) is the maximum real point on the graph.
+130511 I assume that you have this x√(9-x) = x*(9-x)^(1/2)
Taking the first derivative, we have
(9-x)^(1/2) - (1/2)x(9-x)^(-1/2) Setting to 0 and factoring, we have
(9-x)^(-1/2) [ (9-x) - (1/2)x] = 0
[9 - (3/2)x] = 0
9 = (3/2)x
x = 6
We can show that this is a max by taking the second derivative =
[3(x-12)] / [4(9-x)^(3/2)]
And plugging in the critical point x = 6, we see that the numerator < 0 and the denominator > 0 = negative, so this tells us that (6, 6√3) is the maximum real point on the graph.
