Question

Optimize the following function with respect to X, Y=10X^2 – 200X (i.e. find X*, Y*) Then use the second derivate to determine if the point is a maximum or a minimum.

Answer 1

Given,

Y = 10 x^2 - 200x ......equation (1)

differentiating our equation first time we will get,

........equation (2)

putting it equal to 0, we get

10 * 2x - 200 = 0

20x = 200

x =10

Putting the value of X in our equation (1), we can find the value of Y,

Y = 10 * (10)^2 - 200*10

Y = 10 *100 - 2000 = 1000-2000 = -1000

Now, we will do the second derivative test on equation (2) to find out whether our value is maximum or minimum,

Our second derivative value test gives a value of 20 which is greater than 0, hence our value -1,000 is minimum.

Note:

If Value of second derivative test > 0, then the value is minimum.

If Value of second derivative test < 0, then the value is maximum.