Question

Find the minimum value of xy+a^3 (^1⁄x + ^1⁄y).

(a) 3a^2

(b) a^2

(c) a

(d) 1

I had been asked this question in my homework.

The question is from Maxima and Minima of Two Variables topic in chapter Maxima and Minima of Engineering Mathematics

Answer 1

Correct option is (a) 3a^2

For explanation: Given,f(x,y) = \(xy+a^3(\frac{1}{x}+\frac{1}{y})\)

Now, \(\frac{∂f}{∂x}=y-\frac{a^3}{x^2}\) and \(\frac{∂f}{∂y}=x-\frac{a^3}{y^2}\)

Putting, \(\frac{∂f}{∂x}\) and \(\frac{∂f}{∂y}\)=0,and solving two equations,we get,

(x,y)=(a,a) or (-a,a)

Now, at (a,a) r = \(\frac{∂^2 f}{∂x^2}=\frac{2a^3}{x^3}\)=2>0 and \(t=\frac{∂^2 f}{∂y^2}=\frac{2a^3}{y^3}\)=2>0 and \(s=\frac{∂^2 f}{∂x∂y}\)=1

hence, rt-s^2=3>0 and r>0,hence it has minimum value at (a,a).

Now, at (-a,a) r=\(\frac{∂^2 f}{∂x^2}=\frac{2a^3}{x^3}\)=-2<0 and \(t=\frac{∂^2 f}{∂y^2}=\frac{2a^3}{y^3}\)=2>0 and s=\(\frac{∂^2 f}{∂x∂y}\)=1

hence, rt-s^2=-5<0,hence it has no extremum at this point.

Hence maximum value is, f(a,a)=a^2+a^3 \((\frac{1}{a}+\frac{1}{a})=a^2+2a^2=3a^2\)