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Let's solve each of the problems step-by-step:

### 1. Limit Problems
   - **(a)** \(\lim_{x \to 4} \frac{2x - 8}{\sqrt{x} - 2}\)
     - Substitute \( x = 4 \) directly, but since \(\sqrt{4} - 2 = 0\) (resulting in division by zero), we need to simplify. 
     - Factor out the numerator: \(2x - 8 = 2(x - 4)\).
     - Rewrite the limit: 
       \[
       \lim_{x \to 4} \frac{2(x - 4)}{\sqrt{x} - 2}
       \]
     - Rationalize by multiplying the numerator and denominator by the conjugate \(\sqrt{x} + 2\).
     - This should simplify to a value that can be computed by direct substitution.

   - **(b)** \(\lim_{x \to 2} \frac{x^2 + 2x - 8}{x - 2}\)
     - Substitute \( x = 2 \) directly, resulting in a \(0/0\) indeterminate form.
     - Factor the numerator: \(x^2 + 2x - 8 = (x - 2)(x + 4)\).
     - Rewrite the limit as:
       \[
       \lim_{x \to 2} \frac{(x - 2)(x + 4)}{x - 2}
       \]
     - Cancel \(x - 2\) from the numerator and denominator, then substitute \(x = 2\).

### 2. Derivative of \(f(x) = 3x^2 - 6x + 8\) Using the Definition
   - Using the limit definition of the derivative:
     \[
     f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
     \]
   - Substitute \(f(x + h)\) and simplify, then apply the limit as \(h \to 0\).

### 3. First Derivative of Each Function
   - **(a)** \(f(x) = x^5 \left(3\sqrt{2x^4 + 1}\right)\)
     - Use the product rule and chain rule.
   - **(b)** \(f(x) = \cos(x^5 - 2x)\)
     - Use the chain rule.
   - **(c)** \(f(x) = e^{3x^2}\)
     - Use the chain rule.

### 4. Finding the Maximum Value of \(xy\)
   - Given \(x + y = 18\), find the maximum value of \(xy\).
   - Use substitution or Lagrange multipliers to maximize \(xy\) subject to \(x + y = 18\).

### 5. Finding Maximum and Minimum of \(y = x^3 + 3x^2 - 24x\)
   - Take the derivative \(y'\) and set it equal to zero to find critical points.
   - Use the second derivative test or analyze the behavior of \(y'\) around the critical points to determine max/min values.
   - Sketch the graph based on these results.

### 6. Cost and Profit Problem
   - The cost function is given by \(C(x) = 5x^2 - 10x + 30\) (in thousands).
   - If items are sold at \(Rp.50,000.00\) each, find the profit function and optimize for maximum profit.
   - Set up the profit equation, differentiate, and solve for the critical points to find maximum profit.

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### Related Questions
1. How do you simplify a limit using the conjugate method?
2. What is the significance of the derivative definition in finding slopes of tangent lines?
3. How can Lagrange multipliers be used to find constrained maxima and minima?
4. How do we determine the concavity of a function using the second derivative?
5. What role does the cost function play in calculating maximum profit?

### Tip
To tackle optimization problems effectively, always start by setting up equations and identifying constraints. Then, use calculus tools like differentiation to find optimal solutions.

The user submitted a set of calculus and algebra problems, including limits, derivatives, maximum and minimum values, and optimization.

Solve calculus problems involving limits, derivatives, and optimization. Learn how to find maximum and minimum values for functions, use the limit definition of a derivative, and apply critical points for optimization. Ideal for college calculus students.

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