Question

The maximum value of \(x^{4}y^{3}\) such that \(x+y=42\) exists at \(x=\alpha,\; y=\beta\). Then \(\dfrac{\alpha}{\beta}\) in its lowest form is

• A. \(\dfrac{1}{2}\)
• B. \(\dfrac{8}{13}\)
• C. \(3\)
• D. \(\dfrac{1}{6}\)

Hint:

For maximizing \[ x^m y^n \] subject to \[ x+y=k, \] remember directly: \[ x:y=m:n. \] This shortcut saves considerable calculation time.

Answer 1

The Correct Option is C

Concept: When the product of powers of variables is to be maximized subject to a linear constraint, we use the method of differentiation or Lagrange multipliers. For a function of the form \[ x^m y^n \] subject to \[ x+y=k, \] the maximum occurs when \[ \frac{x}{y}=\frac{m}{n}. \] This result follows from logarithmic differentiation.

Step 1: Write the objective function.
We need to maximize \[ f(x,y)=x^4y^3 \] subject to \[ x+y=42. \] Using the constraint, \[ y=42-x. \] Therefore, \[ f(x)=x^4(42-x)^3. \]

Step 2: Take logarithm.
Let \[ z=x^4(42-x)^3. \] Taking logarithm, \[ \ln z = 4\ln x+3\ln(42-x). \] Differentiating with respect to \(x\), \[ \frac{1}{z}\frac{dz}{dx} = \frac{4}{x} -\frac{3}{42-x}. \] For maximum value, \[ \frac{dz}{dx}=0. \] Hence, \[ \frac{4}{x} = \frac{3}{42-x}. \]

Step 3: Solve for \(x\).
Cross multiplying, \[ 4(42-x)=3x. \] \[ 168-4x=3x. \] \[ 168=7x. \] \[ x=24. \] Then \[ y=42-24=18. \]

Step 4: Find the required ratio.
Therefore, \[ \frac{\alpha}{\beta} = \frac{24}{18} = \frac{4}{3}. \] Using the standard result, \[ \frac{x}{y}=\frac{m}{n} = \frac{4}{3}. \] Hence the ratio is \[ \boxed{\frac{4}{3}}. \] Note: There appears to be a discrepancy in the given options. The mathematically correct answer is \[ \boxed{\frac{4}{3}}. \]