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Divide 20 into two parts such that the p...

Divide 20 into two parts such that the product of one part and the cube of the other is maximum. The two parts are

A

`(10,10)`

B

`(5,15)`

C

`(13,7)`

D

none of these

Text Solution

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The correct Answer is:
To solve the problem of dividing 20 into two parts such that the product of one part and the cube of the other is maximized, we can follow these steps: ### Step 1: Define the Variables Let the two parts be \( x \) and \( y \). According to the problem, we have: \[ x + y = 20 \] ### Step 2: Express the Function to Maximize We need to maximize the function \( f(x, y) = x \cdot y^3 \). Since we have a relationship between \( x \) and \( y \), we can express \( x \) in terms of \( y \): \[ x = 20 - y \] Now substitute this into the function: \[ f(y) = (20 - y) \cdot y^3 \] ### Step 3: Simplify the Function Expanding the function gives: \[ f(y) = 20y^3 - y^4 \] ### Step 4: Differentiate the Function To find the maximum, we need to differentiate \( f(y) \) with respect to \( y \): \[ f'(y) = \frac{d}{dy}(20y^3 - y^4) = 60y^2 - 4y^3 \] ### Step 5: Set the Derivative to Zero To find the critical points, set the derivative equal to zero: \[ 60y^2 - 4y^3 = 0 \] Factoring out common terms: \[ 4y^2(15 - y) = 0 \] This gives us: \[ y^2 = 0 \quad \text{or} \quad 15 - y = 0 \] Thus, the critical points are: \[ y = 0 \quad \text{or} \quad y = 15 \] ### Step 6: Determine the Nature of the Critical Points To determine whether these points are maxima or minima, we can use the second derivative test. First, we find the second derivative: \[ f''(y) = \frac{d}{dy}(60y^2 - 4y^3) = 120y - 12y^2 \] Now, evaluate the second derivative at the critical points: 1. For \( y = 0 \): \[ f''(0) = 120(0) - 12(0)^2 = 0 \] (Inconclusive) 2. For \( y = 15 \): \[ f''(15) = 120(15) - 12(15)^2 = 1800 - 2700 = -900 \] (Negative, indicating a maximum) ### Step 7: Find the Corresponding Value of the Other Part Since \( y = 15 \), we can find \( x \): \[ x = 20 - y = 20 - 15 = 5 \] ### Conclusion The two parts are: \[ x = 5 \quad \text{and} \quad y = 15 \] ### Final Answer The two parts are 5 and 15. ---

To solve the problem of dividing 20 into two parts such that the product of one part and the cube of the other is maximized, we can follow these steps: ### Step 1: Define the Variables Let the two parts be \( x \) and \( y \). According to the problem, we have: \[ x + y = 20 \] ### Step 2: Express the Function to Maximize We need to maximize the function \( f(x, y) = x \cdot y^3 \). Since we have a relationship between \( x \) and \( y \), we can express \( x \) in terms of \( y \): ...
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