To find the maximum of a function we need to first find the derivative of it. Equate the derivative to zero and then insert that value in the original function.
But I am unable to use this method on my below functions:
(a) $(100 - q)(3q - 3)$
(b) $2x^2 + 2$, where x must be between 0 and 10
In question (a), i get the equation $-3q^2 + 303q - 300$. Taking the derivative of this we get $-6q + 303$. Equating this to zero we get, $q = \frac{101}{2}$
Then when we substitute $q = \frac{101}{2}$ in $-3q^2 + 303q - 300$ i get some random value.
In question (b), the derivative we get is $4x$. Equating this to zero we get, $4x = 0$ so $x = 0$
. But the question says x must be between 0 and 10.
I am unable to find the maximum of the above functions. where am I going wrong?
I'm going to define $f(q)=(100-q)(3q-3)$.
To find maximum or minimum, we need to differentiate this function, $f'(q)= -3(2q-101)$.
And now, we say $f'(q)=0$. It yields $q=\frac{101}{2}$.
Evaluating this value for $f(q)$, we obtain $f(101/2)=29403/4$.
In the second case, for $2x^2+2$ we can find a global min, at x=0, if the question was focused on only finding the minimum value. In the interval which is given as $x \in [0,10]$, the function evaluated at x=10 will give the maximum value.