Question

Show that the minimum value of U=xy+a^3[1/x+1/y] is 3a^2

Answer 1

Answer:

below is the step by step solution

Step-by-step explanation:

Here the equation is

U = xy + a³×(1/x+1/y) = f(x,y)

Let us suppose b=a³ , then,

f(x,y)= xy + b(1/x+1/y)

df(x,y)= d[xy + b(1/x+1/y)]

df(x,y)= d[xy] +d[ b(1/x+1/y)]

df(x,y)= d[xy] +bd[ (1/x+1/y)]

df(x,y)= xdy+ydx +b[ (-1/x²)dx +(-1/y²)dy]

df(x,y)= xdy+ydx +[ (-b/x²)dx +(-b/y²)dy]

df(x,y)= [y-(b/x²)]dx +[x-(b/y²)]dy

Now, maximum or minimum for x

df(x,y)/dx=0

y-(b/x²)=0

y=b/x²

yx²=b           ...(1)

Now, jaximum or minimum for y

df(x,y)y=0

x-(b/y²)=0

x=b/y²

xy²=b           ...(2)

Here, from equation (1) and (2), we get,

xy² = yx²

xy² - yx²= 0

xy(y-x)=0

x=0 → f(x,y) → ∞ if a>0 and f(x,y) → - ∞ if a<0

or y=0 → f(x,y) → ∞ if a>0 and f(x,y) → - ∞ if a<0

or y=x → f(x,y) = x²+a³(2/x) = x² + (2a³/x)