Math Problem Statement

Biaya untuk memproduksi x bungkus keripik tempe adalah (1/4)x^2 + 25x + 25 ribu rupiah. Jika setiap bungkus keripik dijual dengan harga (55 - 1/2 x) ribu rupiah, maka keuntungan maksimum yang dapat diperoleh adalah....

Solution

The problem you provided involves finding the maximum profit from producing and selling x packs of tempe chips. Here’s a breakdown of the problem:

  1. Cost Function: C(x)=(14x2+25x+25) (in thousand rupiah)C(x) = \left( \frac{1}{4}x^2 + 25x + 25 \right) \text{ (in thousand rupiah)}

  2. Revenue Function: Each pack of tempe chips is sold at the price: P(x)=(5512x) (in thousand rupiah)P(x) = \left( 55 - \frac{1}{2}x \right) \text{ (in thousand rupiah)} So, the total revenue for x packs is: R(x)=x×P(x)=x×(5512x)R(x) = x \times P(x) = x \times \left( 55 - \frac{1}{2}x \right)

  3. Profit Function: Profit is the difference between the revenue and the cost: Profit(x)=R(x)C(x)\text{Profit}(x) = R(x) - C(x)

Let’s calculate the profit function and find its maximum value by determining the critical points. I’ll do the necessary steps now.The maximum profit is obtained when producing and selling 20 packs of tempe chips, and the maximum profit is Rp275,000.00.

Thus, the correct answer is A. Rp275,000.00.

Would you like further details on how the calculations were performed? Here are some related questions you might find helpful:

  1. How do we derive the cost and revenue functions?
  2. What is the significance of taking the derivative in maximizing profit?
  3. How do you solve quadratic equations to find critical points?
  4. What role does concavity play in identifying a maximum?
  5. How can we use similar optimization techniques in real-world business problems?

Tip: For optimization problems like these, remember that maximizing profit often involves taking derivatives of the profit function and solving for critical points where the derivative is zero.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Optimization
Revenue and Cost Analysis

Formulas

Cost Function: C(x) = (1/4)x^2 + 25x + 25
Revenue Function: R(x) = x(55 - 1/2x)
Profit Function: Profit(x) = R(x) - C(x)

Theorems

Maximizing Profit
First Derivative Test for Optimization

Suitable Grade Level

Grades 10-12