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To solve this problem, we need to find two positive numbers \( x \) and \( y \) such that:

1. Their product is 750: \( x \cdot y = 750 \).
2. The sum of one number and ten times the other is minimized: \( S = x + 10y \) or \( S = 10x + y \), depending on which arrangement gives a minimum.

Let's proceed with these steps:

### Step 1: Set Up the Relationship
Let \( x \) and \( y \) be the two positive numbers such that:
\[
x \cdot y = 750
\]
To express \( y \) in terms of \( x \), we can rearrange this equation as:
\[
y = \frac{750}{x}
\]

### Step 2: Set Up the Objective Function
We are asked to minimize the expression for the sum, either \( x + 10y \) or \( 10x + y \). Let's start by assuming the form \( S = x + 10y \).

Substituting \( y = \frac{750}{x} \) into this expression, we get:
\[
S = x + 10\left(\frac{750}{x}\right)
\]
\[
S = x + \frac{7500}{x}
\]

### Step 3: Minimize the Objective Function
To find the minimum of \( S = x + \frac{7500}{x} \), we take the derivative of \( S \) with respect to \( x \) and set it to zero.

#### Find the Derivative
\[
\frac{dS}{dx} = 1 - \frac{7500}{x^2}
\]

#### Set the Derivative to Zero
\[
1 - \frac{7500}{x^2} = 0
\]
\[
\frac{7500}{x^2} = 1
\]
\[
x^2 = 7500
\]
\[
x = \sqrt{7500} = 50\sqrt{3} \approx 86.6
\]

#### Solve for \( y \)
Now, substitute \( x = 86.6 \) back into \( y = \frac{750}{x} \):
\[
y = \frac{750}{86.6} \approx 8.66
\]

### Step 4: Verify Minimum
To confirm that this point gives a minimum, we can check the second derivative or observe the behavior of the function. Given that \( S = x + \frac{7500}{x} \) typically has a minimum for positive values, we can conclude that this choice of \( x \approx 86.6 \) and \( y \approx 8.66 \) minimizes the function.

### Final Answer
The two numbers are approximately:
\[
x \approx 86.6, \quad y \approx 8.66
\]

Would you like further details, or do you have any questions?

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**Related Questions:**
1. How would the result change if the requirement was to maximize instead of minimize?
2. What if the product of the numbers was a different constant, such as 500?
3. How does calculus help in finding minimum values in such problems?
4. Can this problem be solved using different methods besides calculus?
5. What are some real-world applications of minimizing functions with constraints?

**Tip:** In optimization problems, it's essential to differentiate carefully and check whether the critical points give a minimum or maximum based on the problem's requirements.

Find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum.

Explore how to find two positive numbers with a product of 750 that minimizes the sum of one and 10 times the other. Learn step-by-step calculations, involving optimization, derivatives, and calculus concepts. Perfect for high school students in Grades 11-12.

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