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Optimization with constraint (Question #6)

Calculus
Tutor: None Selected Time limit: 0 Hours

Apr 22nd, 2015

As I understand, we speak about question 6.

So, x+y = 2 means y = 2-x, so Q(x) = x^2*(2-x).
x>0 and y>0 means 0<x<2.

Q(0) = 0, Q(2) = 0     (Q(0<x<2) > 0, so it has at least one maximum here)

To find a maximum we ought to find where Q'(x) = 0:
Q'(x) = 2x(2-x) + x^2*(-1) = 4x - 2x^2 - x^2 = 4x - 3x^2 = x*(4 - 3x) = 0.

x1=0 isn't interesting (considered yet), x2 = 4/3. It is belongs to interval (0; 2), so it is (the only) candidate for the point of extremum. The corresponding y is (2 - 4/3) = 2/3.

Nobody asks, but interesting: Q(x2) = (4/3)^2*(2 - 4/3) = (16/9)*(2/3) = 32/27.

Answer: x(max) = 4/3, y(max) = 2/3.

Apr 22nd, 2015

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