91Ó°ÊÓ

In optimization problems, the objective function is a key concept. It is the mathematical expression that you want to maximize or minimize.
The objective function is usually represented by a formula, such as the given equation:
$$ z = 6x - 3x^2 + 2y $$
In this case, we aim to maximize the value of $$ z $$.
We often face other elements like constraints that specify the conditions under which we find this maximum or minimum value.
The goal is to find the value of the decision variables, $$ x $$ and $$ y $$ in this case, that results in the highest value of $$ z $$.

In real-life and mathematical problems, we often encounter limitations or constraints. Constraint optimization involves finding the best solution while satisfying these constraints.
The given constraint here is: $$ y - x^2 = 2 $$
To solve this problem, we need to express $$ y $$ in terms of $$ x $$, using the constraint equation.
First, we rearrange the constraint equation: $$ y = x^2 + 2 $$
Next, substitute $$ y $$ in the objective function:
$$ z = 6x - 3x^2 + 2(x^2 + 2) $$
Simplifying results in
$$ z = 6x - x^2 + 4 $$
By simplifying the function, we make it easier to apply other mathematical techniques to find the optimal solution.

Derivatives help us understand how a function changes as its input changes. They play a crucial role in optimization by identifying the rate of change of the objective function.
To find where a function reaches its maximum or minimum, we need to take its derivative and set it to zero.
In this problem, our simplified objective function in terms of $$ x $$ is:
$$ z = 6x - x^2 + 4 $$
Now, we find the derivative of $$ z $$ with respect to $$ x $$:
$$ \frac{dz}{dx} = 6 - 2x $$
We then set this derivative to zero to find potential values of $$ x $$ that maximize or minimize $$ z $$:
$$ 6 - 2x = 0 $$ -> $$ x = 3 $$
This method ensures that we consider the points where the function's slope is zero, helping us find critical points.

Critical points are where the derivative of a function equals zero, indicating potential maximum, minimum, or saddle points.
For our objective function with the derivative
$$ \frac{dz}{dx} = 6 - 2x $$ -> $$ x = 3 $$
is the critical point.
To confirm if this critical point is a maximum or minimum, we need to check the second derivative.
In this case, the second derivative
$$ \frac{d^2z}{dx^2} = -2 $$
is negative, indicating a maximum.
Once we have $$ x = 3 $$, we use the constraint equation to find the corresponding $$ y $$ value:
$$ y = (3)^2 + 2 = 11 $$
Finally, substituting $$ x = 3 $$ and $$ y = 11 $$ into the objective function:
$$ z = 6(3) - 3(3)^2 + 2(11) = 13 $$
confirms that the maximum value $$ z $$ is 13.