Given that \(xy = \dfrac32\) and both x and y are nonnegative real numbers, find the minimum value of \(10x + \dfrac{3y}5.\)
+118608 Given that \(xy = \dfrac32 \) and both x and y are nonnegative real numbers, find the minimum value of
\(10x + \dfrac{3y}5\)
\(y=\frac{3}{2x}\)
so
\(z=10x+\frac{3y}{5}\\ z=10x+\frac{3}{5}\times \frac{3}{2x}\\ z=10x+\frac{9}{10x}\\ z=10x+0.9x^{-1}\\ \frac{dz}{dx}=10-0.9x^{-2}\\ \frac{d^2z}{dx^2}=1.8x^{-3}>0\qquad \text{So any stationary point will be a minimum.}\\ \text{Find stat point}\\ 10-0.9x^{-2}=0\\ 10=0.9x^{-2}\\ 10=\frac{9}{10x^2}\\ 100x^2=9\\ x^2=\frac{9}{100}\\ x=\frac{3}{10}\qquad \text{Since x>0}\\ so\\ z=10x+0.9x^{-1}\\ min\;\;value\\ =10*\frac{3}{10}+\frac{9*10}{10*3}\\ =3+3\\ =6 \)
check:
Here is the graph:
+118608 Given that \(xy = \dfrac32 \) and both x and y are nonnegative real numbers, find the minimum value of
\(10x + \dfrac{3y}5\)
\(y=\frac{3}{2x}\)
so
\(z=10x+\frac{3y}{5}\\ z=10x+\frac{3}{5}\times \frac{3}{2x}\\ z=10x+\frac{9}{10x}\\ z=10x+0.9x^{-1}\\ \frac{dz}{dx}=10-0.9x^{-2}\\ \frac{d^2z}{dx^2}=1.8x^{-3}>0\qquad \text{So any stationary point will be a minimum.}\\ \text{Find stat point}\\ 10-0.9x^{-2}=0\\ 10=0.9x^{-2}\\ 10=\frac{9}{10x^2}\\ 100x^2=9\\ x^2=\frac{9}{100}\\ x=\frac{3}{10}\qquad \text{Since x>0}\\ so\\ z=10x+0.9x^{-1}\\ min\;\;value\\ =10*\frac{3}{10}+\frac{9*10}{10*3}\\ =3+3\\ =6 \)
check:
Here is the graph:
