Derivatives

1. Convert the inequalities into equations for graphing: - : Rearrange to . - : Rearrange to . - : Rearrange to . 2. Determine the feasible region: - The feasible region is determined by the intersection of the half-planes defined by the inequalities: - : The region above the line . - : The region below the line . - : The region below the line . - : The first quadrant. Graph these constraints and shade the common region that satisfies all the inequalities. 3. Identify the corner points of the feasible region: The corner points are found by solving the intersection points of the boundary lines: - Intersection of and : Solve: From , . Substitute into : Simplify: Substitute into : Corner point: . - Intersection of and : Solve: From , . Substitute into : Substitute into : Corner point: . - Intersection of and : Solve: Subtract from : Substitute into : Corner point: . 4. Evaluate at the corner points: - At : - At : - At : 5. Conclusion: - Maximum value of is at . - Minimum value of is at . Final Answer: - Maximum value: at . - Minimum value: at .

To determine the conditions under which a function is strictly increasing on an interval , we use the following mathematical definition: 1. Strictly Increasing Function: - A function is said to be strictly increasing on an interval if: - This means that the derivative of must be positive throughout the interval. 2. Analysis of Options: - (A): implies is strictly decreasing, not increasing. Hence, incorrect. - (B): correctly implies that is strictly increasing. Hence, correct. - (C): implies that is constant, not strictly increasing. Hence, incorrect. - (D): does not guarantee that is strictly increasing because does not describe the behavior of the derivative. Hence, incorrect. 3. Conclusion: For to be strictly increasing on , the derivative must satisfy: Hence, the correct answer is (B) .

The given function is: To find the derivative of with respect to , we use the chain rule: The derivative of with respect to is: Thus: At , substitute into the expression for : Simplify to : The value of is . Therefore: Hence, the correct answer is (C) .

To find the values of for which is decreasing, we first need to compute the derivative . The derivative of is: For to be decreasing on , we need for all . The expression is bounded, since it is the sum of sinusoidal functions. The maximum value occurs when: Thus, we have the inequality: Solving for , we get: Therefore, the values of for which is decreasing are .

The given function is: To determine for which values of the function is increasing for , we need to find the first derivative of and analyze the condition for increasing functions. The first derivative of is: For the function to be increasing for , we need for all . Thus, we need: Simplifying: For , this inequality will hold true if . Thus, the values of for which is increasing for are .

We are given: and First, differentiate with respect to : Next, differentiate with respect to : To compute , we use the chain rule: Now, substitute the values at . At , , , and . Thus, evaluate: This will give the desired result.

To determine where is strictly increasing, we need to find where its derivative is positive.
Step 1: Find the derivative of We start by differentiating with respect to : Using the chain rule: Factor out the common factor of 3: Step 2: Set the derivative greater than zero to find intervals where is increasing We want : Now, rewrite using the trigonometric identity: So the inequality becomes: Divide both sides by : Step 3: Solve the inequality We know that for , so: Now solve for by first subtracting from each part of the inequality: Now divide by 3: Thus, the function is strictly increasing in the interval . Step 4: Conclude the final intervals Given , the function is strictly increasing in two intervals:

We are given that . First, differentiate with respect to : Using the derivatives and , we get: Next, differentiate to find : Using the derivatives and , we get: Now, we add and : Simplifying: Thus, we have proved that:

To show that the function is increasing, we need to show that the derivative is positive in the interval . First, we differentiate with respect to . Using the chain rule: The derivative of with respect to is , so: Next, we differentiate : Thus, the derivative of is: Now, we need to check the sign of in the interval . In this interval, is positive, since is greater than for . Also, the denominator is always positive. Therefore, in , which implies that is an increasing function on this interval.

We are given that . First, differentiate with respect to : Using the derivatives and , we get: Next, differentiate to find : Using the derivatives and , we get: Now, we add and : Simplifying: Thus, we have proved that:

(i) The total boundary material used includes the perimeter of the rectangular solar panel with an additional partition running parallel to one of the sides. The perimeter of the rectangle is , and the length of the partition is . Therefore, the total boundary material used is: We are given that the total boundary material used is 300 metres, so: (ii) The area of the solar panel is the product of its length and width: From the boundary equation , solve for in terms of : Substitute this expression for into the area equation: Thus, the area of the solar panel as a function of is: (iii) To find the critical points of the area function, we first differentiate with respect to : Set to find the critical points: Now, we check the second derivative to determine whether this critical point is a maximum or minimum: Since , the critical point corresponds to a maximum. Substitute into the equation to find : Thus, the maximum area is: OR
(iii) To calculate the maximum area using the first derivative test, we observe that the first derivative is: For , , indicating that the area is increasing. For , , indicating that the area is decreasing. Therefore, is a maximum, and the maximum area is 3750 square metres, as calculated earlier.

To find the intervals of increase and decrease, we first compute the first derivative of : Now, find the critical points by setting the derivative equal to zero: So, and are the critical points.
To determine the intervals where the function is increasing or decreasing, we analyze the sign of :
- For , (increasing).
- For , (decreasing).
- For , (increasing).
Thus, the function is increasing in and .