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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Application of Derivatives

NDA 2019 · Mathematics · Application of Derivatives

What is the minimum value of a^2 x+b^2 y where x y=c^2 ?

What is the minimum value of a2x+b2ya^2 x+b^2 y where xy=c2x y=c^2 ?

Options

  1. Aabc
  2. B2abc
  3. C3abc
  4. D4abc

Correct answer

B. 2abc

Step-by-step solution

Let p=a2x+b2y\mathrm{p}=\mathrm{a}^2 \mathrm{x}+\mathrm{b}^2 \mathrm{y} and xy=c2⇒y=c2x\mathrm{xy}=\mathrm{c}^2\Rightarrow y=\frac{c^2}{x}...(1) ⇒P=a2x+b2(c2x)Now,dPdx=0⇒a2−b2c2x2=0\Rightarrow P=a^2 x+b^2 \left(\frac{c^2}{x}\right) Now, \frac{d P}{d x}=0 \Rightarrow a²-\frac{b^2 c^2}{x^2}= 0∴y=c2bca=ac2bc=acb⇒a2=b2c2x2⇒x=bca∴Pmin⁡=a2(bca)+b2(acb)=abc+abc=2abc.\begin{aligned}& \therefore y=\frac{c^2}{\frac{b c}{a}}=\frac{a c^2}{b c}=\frac{a c}{b}\\ & \Rightarrow a²=\frac{b^2 c^2}{x^2} \Rightarrow x=\frac{b c}{a}\\ & \therefore P_{\min }=a²\left(\frac{b c}{a}\right)+ b^2\left(\frac{a c}{b}\right)\\ & =\mathrm{abc}+\mathrm{abc}=2 \mathrm{abc} .\end{aligned}$$
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