Math Problem Statement

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

Solution

To solve this problem, we need to find two positive numbers xx and yy such that:

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

Let's proceed with these steps:

Step 1: Set Up the Relationship

Let xx and yy be the two positive numbers such that: xy=750x \cdot y = 750 To express yy in terms of xx, we can rearrange this equation as: y=750xy = \frac{750}{x}

Step 2: Set Up the Objective Function

We are asked to minimize the expression for the sum, either x+10yx + 10y or 10x+y10x + y. Let's start by assuming the form S=x+10yS = x + 10y.

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

Step 3: Minimize the Objective Function

To find the minimum of S=x+7500xS = x + \frac{7500}{x}, we take the derivative of SS with respect to xx and set it to zero.

Find the Derivative

dSdx=17500x2\frac{dS}{dx} = 1 - \frac{7500}{x^2}

Set the Derivative to Zero

17500x2=01 - \frac{7500}{x^2} = 0 7500x2=1\frac{7500}{x^2} = 1 x2=7500x^2 = 7500 x=7500=50386.6x = \sqrt{7500} = 50\sqrt{3} \approx 86.6

Solve for yy

Now, substitute x=86.6x = 86.6 back into y=750xy = \frac{750}{x}: y=75086.68.66y = \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+7500xS = x + \frac{7500}{x} typically has a minimum for positive values, we can conclude that this choice of x86.6x \approx 86.6 and y8.66y \approx 8.66 minimizes the function.

Final Answer

The two numbers are approximately: x86.6,y8.66x \approx 86.6, \quad y \approx 8.66

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


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.

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Math Problem Analysis

Mathematical Concepts

Optimization
Product and Sum Relationships
Derivatives
Calculus

Formulas

Product constraint: x * y = 750
Objective function for minimization: S = x + 10y
Substitution for y: y = 750 / x
Derivative to find minimum: dS/dx = 1 - 7500 / x^2

Theorems

Optimization using Calculus
First Derivative Test

Suitable Grade Level

Grades 11-12