Consider the following for the next items that follow: Given that 4x2 + y2 = 9.
What is the maximum value of xy ?
Finding the Maximum Value of xy Given 4x² + y² = 9
The problem asks for the maximum possible value of the product \(xy\), given the constraint equation \(4x^2 + y^2 = 9\).
We are given the equation: \[4x^2 + y^2 = 9\] We need to find the maximum value of \(xy\).
Using the AM-GM Inequality to Find Maximum xy
A common method to find the maximum or minimum value of an expression involving terms whose sum is constant (or related) is using the Arithmetic Mean-Geometric Mean (AM-GM) inequality. The AM-GM inequality states that for any non-negative real numbers \(a\) and \(b\), the arithmetic mean is greater than or equal to the geometric mean:
\[\frac{a+b}{2} \ge \sqrt{ab}\]Equality holds if and only if \(a = b\).
In our given equation \(4x^2 + y^2 = 9\), the terms \(4x^2\) and \(y^2\) are squares of real numbers (multiplied by a positive constant), so they are always non-negative for real values of \(x\) and \(y\). Thus, we can apply the AM-GM inequality to these two terms, \(a = 4x^2\) and \(b = y^2\).
Applying the inequality:
\[\frac{4x^2 + y^2}{2} \ge \sqrt{(4x^2)(y^2)}\]We know from the given equation that \(4x^2 + y^2 = 9\). Substituting this value into the inequality:
\[\frac{9}{2} \ge \sqrt{4x^2y^2}\]Simplify the right side of the inequality:
\[\frac{9}{2} \ge \sqrt{(2xy)^2}\] \[\frac{9}{2} \ge |2xy|\]The expression \(|2xy|\) represents the absolute value of \(2xy\). The inequality \(\frac{9}{2} \ge |2xy|\) means that \(|2xy|\) must be less than or equal to \(\frac{9}{2}\). This can be written as:
\[-\frac{9}{2} \le 2xy \le \frac{9}{2}\]To find the range for \(xy\), we divide all parts of the inequality by 2:
\[-\frac{9}{4} \le xy \le \frac{9}{4}\]This inequality tells us that the value of \(xy\) can range from \(-\frac{9}{4}\) to \(\frac{9}{4}\). Therefore, the maximum value that \(xy\) can attain is \(\frac{9}{4}\).
The maximum value is achieved when the equality holds in the AM-GM inequality, which is when \(4x^2 = y^2\). Let's verify this condition using the original equation \(4x^2 + y^2 = 9\).
If \(4x^2 = y^2\), substitute \(y^2 = 4x^2\) into the original equation:
\[4x^2 + (4x^2) = 9\] \[8x^2 = 9\] \[x^2 = \frac{9}{8}\] \[x = \pm\sqrt{\frac{9}{8}} = \pm\frac{3}{\sqrt{8}} = \pm\frac{3}{2\sqrt{2}} = \pm\frac{3\sqrt{2}}{4}\]Since \(y^2 = 4x^2\), we have \(y = \pm\sqrt{4x^2} = \pm 2|x|\). If \(x = \frac{3\sqrt{2}}{4}\), then \(|x| = \frac{3\sqrt{2}}{4}\), so \(y = \pm 2(\frac{3\sqrt{2}}{4}) = \pm \frac{3\sqrt{2}}{2}\). If \(x = -\frac{3\sqrt{2}}{4}\), then \(|x| = \frac{3\sqrt{2}}{4}\), so \(y = \pm 2(\frac{3\sqrt{2}}{4}) = \pm \frac{3\sqrt{2}}{2}\).
Let's check the value of \(xy\) when equality holds (\(y^2 = 4x^2\), which implies \(y = \pm 2x\)).
If \(y = 2x\), then \(xy = x(2x) = 2x^2\). Substitute \(x^2 = \frac{9}{8}\): \(xy = 2\left(\frac{9}{8}\right) = \frac{18}{8} = \frac{9}{4}\)
If \(y = -2x\), then \(xy = x(-2x) = -2x^2\). Substitute \(x^2 = \frac{9}{8}\): \(xy = -2\left(\frac{9}{8}\right) = -\frac{18}{8} = -\frac{9}{4}\)
The maximum value of \(xy\) is indeed \(\frac{9}{4}\).
Summary of the Steps
- Start with the given equation \(4x^2 + y^2 = 9\).
- Recognize that \(4x^2\) and \(y^2\) are non-negative terms.
- Apply the AM-GM inequality to \(4x^2\) and \(y^2\): \(\frac{4x^2 + y^2}{2} \ge \sqrt{(4x^2)(y^2)}\).
- Substitute \(4x^2 + y^2 = 9\) into the inequality: \(\frac{9}{2} \ge \sqrt{4x^2y^2}\).
- Simplify to get \(\frac{9}{2} \ge |2xy|\).
- Rewrite the absolute value inequality as \(-\frac{9}{2} \le 2xy \le \frac{9}{2}\).
- Divide by 2 to find the range of \(xy\): \(-\frac{9}{4} \le xy \le \frac{9}{4}\).
- Identify the maximum value from this range.
The maximum value of \(xy\) is \(\frac{9}{4}\).
| Concept | Description |
|---|---|
| AM-GM Inequality | For non-negative \(a, b\), \(\frac{a+b}{2} \ge \sqrt{ab}\). Equality holds if \(a=b\). |
| Given Equation | \(4x^2 + y^2 = 9\) |
| Terms for AM-GM | \(a = 4x^2, b = y^2\) (both non-negative) |
| Applied AM-GM | \(\frac{4x^2+y^2}{2} \ge \sqrt{4x^2 y^2}\) |
| Using Given | \(\frac{9}{2} \ge \sqrt{(2xy)^2} \implies \frac{9}{2} \ge |2xy|\) |
| Inequality for xy | \(-\frac{9}{4} \le xy \le \frac{9}{4}\) |
| Maximum Value | \(\frac{9}{4}\) |
Revision Table: Key Concepts for Optimization Problems
| Topic | Relevance | Notes |
|---|---|---|
| AM-GM Inequality | Useful for finding min/max of products or sums of non-negative terms. | Requires terms to be non-negative. Equality condition is key for extremum. |
| Calculus (Derivatives) | Can be used to find extrema of functions. | Could express \(y\) in terms of \(x\) (or vice versa), substitute into \(xy\), and differentiate. Requires careful handling of domain. |
| Lagrange Multipliers | Method for finding extrema of a function subject to constraints. | More advanced technique for optimizing \(f(x,y)=xy\) subject to \(g(x,y)=4x^2+y^2-9=0\). |
| Quadratic Forms | Understanding the geometry of the constraint (an ellipse). | The equation \(4x^2 + y^2 = 9\) represents an ellipse centered at the origin. Finding max/min of \(xy\) on this ellipse. |
Additional Information: Alternative Approaches to Maximize xy
While AM-GM is efficient here, other methods could also be used to solve this optimization problem:
- Calculus Approach: You could express \(y\) in terms of \(x\) from the constraint, \(y^2 = 9 - 4x^2\). Since we are considering real numbers, \(9 - 4x^2 \ge 0\), which means \(4x^2 \le 9\), or \(x^2 \le \frac{9}{4}\). Thus, \(-\frac{3}{2} \le x \le \frac{3}{2}\). Then \(y = \pm\sqrt{9 - 4x^2}\). We want to maximize \(f(x) = x \cdot (\pm\sqrt{9 - 4x^2})\). This involves considering two cases (\(y = \sqrt{9 - 4x^2}\) and \(y = -\sqrt{9 - 4x^2}\)) and differentiating \(x\sqrt{9 - 4x^2}\) or \(-x\sqrt{9 - 4x^2}\) with respect to \(x\) to find critical points. This method is generally more complex algebraically than AM-GM for this specific problem.
- Trigonometric Substitution: The equation \(4x^2 + y^2 = 9\) can be rewritten as \((\frac{x}{3/2})^2 + (\frac{y}{3})^2 = 1\). This suggests a substitution like \(x = \frac{3}{2}\cos\theta\) and \(y = 3\sin\theta\). Then \(xy = (\frac{3}{2}\cos\theta)(3\sin\theta) = \frac{9}{2}\cos\theta\sin\theta = \frac{9}{4}(2\cos\theta\sin\theta) = \frac{9}{4}\sin(2\theta)\). The maximum value of \(\sin(2\theta)\) is 1. Therefore, the maximum value of \(xy\) is \(\frac{9}{4} \times 1 = \frac{9}{4}\). This method is also quite elegant for this type of constraint.
Both calculus and trigonometric substitution confirm the maximum value found using the AM-GM inequality.
Important Questions from Applications of Derivatives
The function is decreasing on :
The function attains local minimum value at :
What is the maximum value of y?
Consider the following statements:
1. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} + {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on [0, ∞).
2. \({\rm{y}} = \frac{{{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{ - {\rm{x}}}}}}{2}\) is an increasing function on (-∞, ∞).
Which of the above statements is/are correct?
\({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 1}}{{{{\rm{x}}^2} + 1}}\) where x ϵ R
At what value of x does f(x) attain minimum value?
