91Ó°ÊÓ

A quadratic equation is a type of algebraic expression that takes the form \(ax^2 + bx + c = 0\). In our context, the expression appears as part of a more complicated equation:

• \(4x^2 + 8xy + 9y^2 - 8x - 24y + 4 = 0\).

This quadratic equation involves two variables, \(x\) and \(y\), forming a curve in a two-dimensional plane. Solving this equation helps us find specific points where certain conditions, such as a derivative equaling zero, are met. When you encounter a quadratic equation, one useful method of solving for one variable, such as \(y\), is through substitution or by using the quadratic formula:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In this scenario, the quadratic is rearranged in terms of \(y\) and solved, which helps in finding the maximum and minimum values of \(y\) under specific conditions. Finding such values involves calculating roots that satisfy the equation, a crucial technique in various fields of mathematics.

Critical points are where the first derivative of a function is zero or undefined. These include points where a curve changes direction, which could be peaks (maximums), valleys (minimums), or saddle points. In this problem, finding where \(\frac{dy}{dx} = 0\) helps identify these critical points. Firstly, we implicitly differentiate the entire equation:

• \(4x^2 + 8xy + 9y^2 - 8x - 24y + 4 = 0\).

Once implicit differentiation is applied, an expression for \(\frac{dy}{dx}\) is derived. To locate critical points, set \(\frac{dy}{dx}=0\) and solve:

• \(x + y = 1\).

This shows there’s a shared linear relationship between \(x\) and \(y\) at these points. Critical points offer significant insight into the geometric nature of a problem, helping identify where local maximums or minimums occur on a graph.

The second derivative test assists in determining the nature of critical points found with the first derivative test. By analyzing changes in concavity, it identifies whether a point is a local maximum, minimum, or possibly a point of inflection. After finding \(\frac{dy}{dx}\), we apply a further differentiation step to obtain \(\frac{d^2y}{dx^2}\). This is calculated by differentiating the expression for \(\frac{dy}{dx}\), using the quotient rule for differentiation due to its fractional form. The test checks the sign of \(\frac{d^2y}{dx^2}\) at each critical point:

• If \(\frac{d^2y}{dx^2} > 0\), the point is a local minimum.
• If \(\frac{d^2y}{dx^2} < 0\), the point is a local maximum.
• If \(\frac{d^2y}{dx^2} = 0\), the test is inconclusive.

In the exercise, we validate that at critical points determined earlier, the second derivative matches the given expression \(\frac{4}{8 - 5y}\). This ensures we correctly identify whether these are points of maxima or minima, as exemplified by finding \(y\)'s maximum and minimum values are \(\frac{2}{3}\) and \(-\frac{1}{3}\) respectively.