If xy² = 27, find minimum value of 32x + y where x and y > 0.

We are tasked with minimizing \( 32x + y \), given that \( xy^2 = 27 \) and both \( x \) and \( y \) are greater than zero.

### Step 1: Express \( x \) in terms of \( y \)
From the given equation:
\[
xy^2 = 27 \implies x = \frac{27}{y^2}
\]
Now, substitute this expression for \( x \) into the objective function \( 32x + y \):
\[
32x + y = 32\left(\frac{27}{y^2}\right) + y = \frac{864}{y^2} + y
\]

### Step 2: Minimize the function
We now minimize the function \( f(y) = \frac{864}{y^2} + y \). To find the critical points, take the derivative of \( f(y) \) with respect to \( y \) and set it equal to zero:
\[
f'(y) = \frac{d}{dy} \left( \frac{864}{y^2} + y \right) = -\frac{1728}{y^3} + 1
\]
Set this equal to zero to find the critical points:
\[
-\frac{1728}{y^3} + 1 = 0 \implies \frac{1728}{y^3} = 1 \implies y^3 = 1728
\]
Taking the cube root of both sides:
\[
y = 12
\]

### Step 3: Find \( x \) when \( y = 12 \)
Now substitute \( y = 12 \) back into the equation \( xy^2 = 27 \) to find \( x \):
\[
x = \frac{27}{y^2} = \frac{27}{12^2} = \frac{27}{144} = \frac{3}{16}
\]

### Step 4: Compute the minimum value of \( 32x + y \)
Now substitute \( x = \frac{3}{16} \) and \( y = 12 \) into the objective function:
\[
32x + y = 32\left(\frac{3}{16}\right) + 12 = 6 + 12 = 18
\]

Thus, the minimum value of \( 32x + y \) is \( \boxed{18} \).

Would you like more details on any step, or have any other questions?

Here are 5 related questions:
1. What is the significance of using derivatives to find minimum values?
2. How do you find critical points in an optimization problem?
3. What other methods can be used to solve such optimization problems?
4. Can the Lagrange multiplier method be applied to this problem?
5. How does changing the constraint \( xy^2 = 27 \) affect the solution?

**Tip:** Always check the second derivative when finding minima or maxima to ensure it's a minimum or maximum point.

If xy² = 27, find the minimum value of 32x + y where x and y > 0.

Find the minimum value of 32x + y under the constraint xy² = 27, using derivatives for optimization. This problem covers key algebraic and calculus concepts, including derivative tests and optimization techniques. Suitable for Grades 11-12 students.

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