Application Of Derivatives

Take log: Differentiate w.r.t. : Collect: So,

To show that the area is maximum at the critical point, we take the second derivative of with respect to and check its sign at . The second derivative is given by: This will involve applying the quotient rule and simplifying. For brevity, the details can be carried out to find that the second derivative is negative, indicating a maximum at .

We are given the function: To determine where the function is increasing or decreasing, we need to find the first derivative of with respect to . Using the quotient rule: where and . Differentiating and : Now, applying the quotient rule: Since the derivative is always positive (as for all ), the function is always increasing for . Thus, is an increasing function for all values of except .

Given: - The length of the ladder ,
- The rate of change of distance is ,
- The foot of the ladder is 3 m away from the wall, i.e., .
From the Pythagorean theorem: Differentiating with respect to time : Substitute the given values: Now, solve for : To find , use the Pythagorean theorem: Thus: So, the height on the wall is increasing at a rate of when the foot of the ladder is 3 m away from the wall.

To find the derivative of the area with respect to , we apply the product and chain rule: Differentiating with respect to : Now, using the chain rule on the second term: Thus: Simplifying: To find the critical point, set : So, the critical point occurs when .

Let the length of the ladder be , the height of the ladder on the wall be , and the distance of the foot of the ladder from the wall be . Using the Pythagorean theorem, we have: From equation (1), express in terms of and : Next, the area of the right triangle formed by the ladder, wall, and ground is given by:

The slope of the curve at any point is given by the first derivative , which represents the rate of change of with respect to . To find the maximum slope, we need to maximize . Step 1: Compute the First Derivative
Differentiate with respect to : Step 2: Find the Maximum Value of the Slope
To maximize , we take the second derivative and find the critical points where the slope of is zero (i.e., where has a maximum or minimum). Differentiate with respect to : Set the second derivative equal to zero to find critical points: Step 3: Determine the Nature of the Critical Point
Use the second derivative test to check if is a maximum: - If , the function has a local maximum. - If , the function has a local minimum. Substitute into : Since at , the second derivative test is inconclusive. We need to check the first derivative around or use the first derivative test. Step 4: First Derivative Test
Evaluate around : - For : - For : - For : The calculation seems inconsistent. Let's compute correctly at : Let's try points closer to check the behavior: - For : - For : This indicates a minimum, not a maximum. Let's find the critical point by setting the first derivative to zero:
Divide by 3: Discriminant: Since the discriminant is negative, has no real roots, meaning for all (as the quadratic opens upwards and the minimum value is positive). Step 5: Re-evaluate the Maximum Slope
Since is a quadratic in with a positive leading coefficient, it has a minimum value (not a maximum) at its vertex. The vertex of occurs at : At : As , .
Let's check the behavior:
- For : - For : The slope increases without bound as increases. However, the problem asks for the maximum slope, which suggests a possible misinterpretation. Let's assume the intent is to find the maximum value of the slope function within a practical context or correct the function. The slope has no upper bound, but let's derive correctly.

Step 1: Differentiate the function.
Given
Derivative:

Step 2: Find critical points.
Set derivative equal to zero:



Step 3: Use second derivative test.
Second derivative:

Check at :
Therefore gives local minimum.
Value:

Check at :
Therefore gives local maximum.
Value:

Step 4: Conclusion.
Local minimum value =
Local maximum value =
Hence the correct statement is:

Final Answer:

Step 1: Define the variables.
Let the length of the rectangle be and the breadth be .
Given that the perimeter of the rectangle is When the rectangle is revolved about one of its sides, it generates a cylinder.
In this cylinder: Step 2: Write the formula for volume of the cylinder.
The volume of a cylinder is Substituting and : Using : Step 3: Express the volume as a function of .
Expand the expression: Thus Step 4: Differentiate to find maximum volume.
For maximum volume Divide by : Thus But if , then , which is not possible.
Hence Step 5: Find the breadth.
Step 6: Verify maximum condition.
The second derivative Substitute : Hence the volume is maximum.
Step 7: Final dimensions.
Thus the rectangle must have sides cm and cm to produce the cylinder of maximum volume.
Final Answer: