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.

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