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The problem in the image involves determining the optimal price for maximizing profits when adjusting the price and sales of computers. To solve this, we need to define two variables and form a system of equations.

### Problem Breakdown:
1. The initial price per computer is 22 million VND, with projected sales of 500 units per year.
2. For every 200,000 VND reduction in the price, sales increase by 50 units.

### Let’s define the two variables:
- Let \( x \) be the number of price reductions in units of 200,000 VND.
- Let \( P \) be the new price per computer after the price reductions.

The total sales volume will be \( 500 + 50x \), since each price reduction increases sales by 50 units.

The new price \( P \) will be \( 22 - 0.2x \) million VND, as each price reduction is 0.2 million VND.

### Form the profit function:
The total profit can be represented as:

\[
\text{Profit} = (\text{price per unit}) \times (\text{number of units sold})
\]

This gives:

\[
\text{Profit} = (22 - 0.2x) \times (500 + 50x)
\]

### Now, we’ll expand this equation and maximize it with respect to \( x \).

Let’s calculate the expression for profit.The profit function is given by:

\[
\text{Profit} = -10x^2 + 1000x + 11000
\]

This is a quadratic function of \( x \), which represents the number of price reductions. To find the value of \( x \) that maximizes the profit, we need to find the vertex of this quadratic function.

The formula to find the \( x \)-value of the vertex is:

\[
x = -\frac{b}{2a}
\]

where \( a = -10 \) and \( b = 1000 \).

Let’s calculate the optimal value of \( x \).The optimal number of price reductions is \( x = 50 \). This means the price should be reduced by:

\[
0.2 \times 50 = 10 \text{ million VND}
\]

### Final price:
\[
\text{New price per computer} = 22 - 10 = 12 \text{ million VND}
\]

At this price, the profit will be maximized.

---

Would you like further details or any clarifications? Here are five related questions you can explore:

1. How would a further reduction in price affect profits?
2. How is the maximum profit calculated using this result?
3. What is the total number of computers sold at the optimal price?
4. Can we graph the profit function to visually confirm the result?
5. How does the price elasticity of demand factor into this solution?

**Tip:** Quadratic equations are often used to find maximum or minimum values in optimization problems, such as maximizing profit.

Cửa hàng nhà bác Dũng chuyên kinh doanh máy tính tại Hà Nội. Một loại máy tính có giá nhập vào một chiếc là 18 triệu đồng và bán ra với giá 22 triệu đồng. Với giá bán như trên thì một năm số lượng máy tính bán được dự kiến là 500 chiếc. Để tăng thêm lượng tiêu thụ dòng máy tính này, bác Dũng dự định giảm giá bán và ước lượng cứ giảm 200 nghìn đồng một chiếc thì số lượng máy tính bán ra trong một năm sẽ tăng thêm 50 chiếc. Vậy bác Dũng phải bán với giá bao nhiêu để sau khi giảm giá lợi nhuận thu được sẽ cao nhất?

Learn how to maximize profit by adjusting the price and sales of computers. This problem involves optimization using quadratic functions, and helps calculate the optimal price reduction to increase sales and maximize profit. Suitable for Grades 10-12 students.

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