Problem 51
Question
If \(a\) and \(b\) are positive numbers, find the maximum value of \(f(x)=x^{a}(1-x)^{b}, 0 \leqslant x \leqslant 1\)
Step-by-Step Solution
Verified Answer
The maximum value is \(\left(\frac{a}{a+b}\right)^a\left(\frac{b}{a+b}\right)^b\).
1Step 1: Find the derivative
First, we need to calculate the derivative of the function \(f(x)=x^a(1-x)^b\). We use the product rule of differentiation. Let \(u = x^a\) and \(v = (1-x)^b\). The derivatives are \(u' = ax^{a-1}\) and \(v' = -b(1-x)^{b-1}\). Applying the product rule: \[f'(x) = u'v + uv' = ax^{a-1}(1-x)^b - bx^a(1-x)^{b-1}\]
2Step 2: Simplify the derivative
Factor common terms from the derivative: \[f'(x) = x^{a-1}(1-x)^{b-1} [a(1-x) - bx]\]This simplifies to: \[f'(x) = x^{a-1}(1-x)^{b-1} [a - ax - bx]\]\[f'(x) = x^{a-1}(1-x)^{b-1} (a - x(a+b))\]
3Step 3: Set derivative equal to zero
To find critical points, set the derivative equal to zero: \[x^{a-1}(1-x)^{b-1} (a - x(a+b)) = 0\]Since \(x^{a-1}(1-x)^{b-1} eq 0\), solve \(a - x(a+b) = 0\).
4Step 4: Solve for x
Rearrange the equation \(a - x(a+b) = 0\) to solve for \(x\): \[x(a+b) = a\]\[x = \frac{a}{a+b}\]
5Step 5: Check endpoints and plug in critical point
Evaluate \(f(x)\) at \(x = 0, x = 1\), and the critical point \(x = \frac{a}{a+b}\):- \(f(0) = 0^a(1-0)^b = 0\)- \(f(1) = 1^a(1-1)^b = 0\)- \(f\left(\frac{a}{a+b}\right) = \left(\frac{a}{a+b}\right)^a\left(1-\frac{a}{a+b}\right)^b = \left(\frac{a}{a+b}\right)^a\left(\frac{b}{a+b}\right)^b\)
6Step 6: Conclude the maximum value
Since \(f(x)\) at the endpoints is 0, the maximum value is achieved at \(x = \frac{a}{a+b}\). Therefore, the maximum value is: \[\left(\frac{a}{a+b}\right)^a\left(\frac{b}{a+b}\right)^b\]
Key Concepts
Critical PointsProduct RuleDerivative SimplificationMaximum and Minimum Values
Critical Points
Critical points in calculus represent the values from the domain of a function where the first derivative is zero or undefined.
These points are important because they help to locate potential maximum or minimum values of the function.
When finding the critical points for a function, we start by taking the derivative and setting it equal to zero. For example, with the function given by \(f(x)=x^a(1-x)^b\), the derivative \((f'(x))\) is calculated, and we identify when \(f'(x) = 0\).
In our case, after simplifying, we end up with the expression \(x^{a-1}(1-x)^{b-1} (a-x(a+b)) = 0\).
This factorization helps us see that to find the critical points, we need to solve \(a - x(a+b) = 0\), which leads to the critical point \(x = \frac{a}{a+b}\).
Whenever you do this in practice, focus on painstakingly solving every term that can contribute to zero in your equation!
These points are important because they help to locate potential maximum or minimum values of the function.
When finding the critical points for a function, we start by taking the derivative and setting it equal to zero. For example, with the function given by \(f(x)=x^a(1-x)^b\), the derivative \((f'(x))\) is calculated, and we identify when \(f'(x) = 0\).
In our case, after simplifying, we end up with the expression \(x^{a-1}(1-x)^{b-1} (a-x(a+b)) = 0\).
This factorization helps us see that to find the critical points, we need to solve \(a - x(a+b) = 0\), which leads to the critical point \(x = \frac{a}{a+b}\).
Whenever you do this in practice, focus on painstakingly solving every term that can contribute to zero in your equation!
Product Rule
The product rule is a fundamental differentiation technique used when a function is the product of two or more functions.
It's essential because not all functions are simple sums or powers; often, they're products as we see with \(f(x)=x^a(1-x)^b\).
The rule states that if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by:
Understanding the product rule helps in breaking down and managing complex derivatives in calculus problems.
It's essential because not all functions are simple sums or powers; often, they're products as we see with \(f(x)=x^a(1-x)^b\).
The rule states that if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by:
- \(f'(x) = u'(x)v(x) + u(x)v'(x)\)
Understanding the product rule helps in breaking down and managing complex derivatives in calculus problems.
Derivative Simplification
Once we've applied rules like the product rule, our resulting derivative expressions often become quite complex.
Simplification becomes an effective tool in solving these derivatives, as it helps in easily spotting solutions for a given problem.
Take, for instance, the derivative from our exercise:\[f'(x) = ax^{a-1}(1-x)^b - bx^a(1-x)^{b-1}\]To simplify, common terms \(x^{a-1}(1-x)^{b-1}\) are factored out:\[f'(x) = x^{a-1}(1-x)^{b-1} [a(1-x) - bx]\]This simplifies further to:\[f'(x) = x^{a-1}(1-x)^{b-1} (a - x(a+b))\]The key here is identifying and factoring out common terms so that solving for critical points becomes more straightforward. Simplifying derivatives not only provides clarity but also reduces errors in finding solutions.
Simplification becomes an effective tool in solving these derivatives, as it helps in easily spotting solutions for a given problem.
Take, for instance, the derivative from our exercise:\[f'(x) = ax^{a-1}(1-x)^b - bx^a(1-x)^{b-1}\]To simplify, common terms \(x^{a-1}(1-x)^{b-1}\) are factored out:\[f'(x) = x^{a-1}(1-x)^{b-1} [a(1-x) - bx]\]This simplifies further to:\[f'(x) = x^{a-1}(1-x)^{b-1} (a - x(a+b))\]The key here is identifying and factoring out common terms so that solving for critical points becomes more straightforward. Simplifying derivatives not only provides clarity but also reduces errors in finding solutions.
Maximum and Minimum Values
Calculating maximum and minimum values is one of the end goals of employing calculus techniques.
These values help in understanding the behavior of functions within a given interval, answering real-world problems like optimization.
After finding the critical points, like \(x = \frac{a}{a+b}\) for our function, determine which points give maximum or minimum values by comparing \(f(x)\) at critical points and endpoints.
With \(f(x)=x^a(1-x)^b\), check the function at \(x = 0\), \(x = 1\), and \(x = \frac{a}{a+b}\):
Mastering this concept allows you to identify optimal conditions in various scenarios!
These values help in understanding the behavior of functions within a given interval, answering real-world problems like optimization.
After finding the critical points, like \(x = \frac{a}{a+b}\) for our function, determine which points give maximum or minimum values by comparing \(f(x)\) at critical points and endpoints.
With \(f(x)=x^a(1-x)^b\), check the function at \(x = 0\), \(x = 1\), and \(x = \frac{a}{a+b}\):
- \(f(0) = 0\)
- \(f(1) = 0\)
- \(f\left(\frac{a}{a+b}\right) = \left(\frac{a}{a+b}\right)^a\left(\frac{b}{a+b}\right)^b\)
Mastering this concept allows you to identify optimal conditions in various scenarios!
Other exercises in this chapter
Problem 50
The family of bell-shaped curves $$y=\frac{1}{\sigma \sqrt{2 \pi}} e^{-(x-\mu)^{2} /\left(2 \sigma^{2}\right)}$$ occurs in probability and statistics, where it
View solution Problem 51
What constant acceleration is required to increase the speed of a car from 30 \(\mathrm{mi} / \mathrm{h}\) to 50 \(\mathrm{mi} / \mathrm{h}\) in 5 \(\mathrm{s}
View solution Problem 51
\(49-52=\) The line \(y=m x+b\) is called a slant asymptote if \(f(x)-(m x+b) \rightarrow 0\) as \(x \rightarrow \infty\) or \(x \rightarrow-\infty\) because th
View solution Problem 51
Let \(v_{1}\) be the velocity of light in air and \(v_{2}\) the velocity of light in water. According to Fermat's Principle, a ray of light will travel from a p
View solution