fxyxy2 subject to the constraint 1x7y21 Max valueSolutionAs

Solution

As we are not told that we must use LaGrange multipliers, the most straightforward thing to do is, as

x + 7y² = 1, substitute x = 1 -  7y² into

f(x,y) = xy².

We thus get f(x,y) = xy² = (1 -  7y²)y² = y² - 7y⁴

FYI, as this equals y²(1 - 7y²), we know that this function has a relative minimum at 0 and 2 global maxima in the interval (-1/7, 0) and (0, 1/7)

Taking the derivative, 2y - 28y³ = 0.

2y(1 - 14y²) = 0

y = 0, -1/14, 1/14 or 0,-14/14,14/14

Taking the second derivative, we get 2 - 84y²

At y = 0,  2 - 84y² = 2 is positive, so this is a relative minimum (as stated above)

At y = ±14/14,  2 - 84y² = 2 - 84(1/14) = 2 - 6 = - 4 < 0, so there are maxima at y = ±14/14

When y = ±14/14, x = 1 - 7y² = 1 - 7(1/14) = 1/2

Thus, our maxima are at (1/2, -14/14) and (1/2, 14/14)

At both of the maxima, f(x,y) =

xy² = 1/2(1/14) = 1/28

Thus, (1/2, -14/14) (1/2, 14/14) have the same global maximum, 1/28

(Of course, the local minimum, when y = 0, will have  x = 1 -  7y² = 1 - 0 = 1 and f(x,y) = xy² = 0;

Thus, the local minimum, at (1, 0), = 0

There is no global minimum, as f(x,y) goes to - as y goes to ± and x goes to -