Class 12 Applications Of Derivatives · Very Long Answer5 Marksconstrained_optimization
Class 12 Maths · Applications Of Derivatives · Very Long Answer #58
For positive constants a,b,c and positive x,y with xy=c², find the minimum of ax+by.
✅ Step-by-Step Solution
Final Answer: 2c√(ab), at x=c√(b/a), y=c√(a/b).
- From xy=c² obtain y=c²/x and F(x)=ax+bc²/x, x>0.
- Differentiate: F′=a−bc²/x².
- The unique positive stationary point is x₀=c√(b/a).
- F′<0 below x₀ and F′>0 above x₀, proving the global minimum; y₀=c√(a/b).
- Substitution gives Fmin=ac√(b/a)+bc√(a/b)=2c√(ab).
📋 CBSE Stepwise Marking Scheme
- [1 mark] From xy=c² obtain y=c²/x and F(x)=ax+bc²/x, x>0.
- [1 mark] Differentiate: F′=a−bc²/x².
- [1 mark] The unique positive stationary point is x₀=c√(b/a).
- [1 mark] F′<0 below x₀ and F′>0 above x₀, proving the global minimum; y₀=c√(a/b).
- [1 mark] Substitution gives Fmin=ac√(b/a)+bc√(a/b)=2c√(ab).
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