Given that $xy = \dfrac32$ and both $x$ and $y$ are nonnegative real numbers, find the minimum value of $10x + \dfrac{3y}5.$
+128448 \(xy = \dfrac32 \) \(10x + \dfrac{3y}5 \)
y = 3/[2x]
So....let f(x) = 10x + 3 [ 3]/ [ 5 * 2x] = 10x + 9/[10x ] = 10x + (9/10)x^-1
Take the derivative , set to 0 and solve for x
10 - (9/10)x^-2 = 0
10 = (9/10)x^-2
10x^2 = 9/10
x^2 = 9/100
x = 3/10
And we can prove that x = 3/10 minimizes f(x) by evaluating f " (3/10)
2(9/10)
_______ > 0 so x = 3/10 minimizes f(x)
(3/10)^3
And y = 3 /[2(3/10)] = 3 / (6/10) = 30/6 = 5
And f(x) = 10(3/10) + 3(5)/5 = 3 + 3 = 6 = minimum for f(x)
