91Ó°ÊÓ

The given functions are$2x+4y=5$ and$x^2+z^2=2y$ for which maximum z are required.

Lagrange multipliers can be utilised to find the maximum and minimum of any problem.

Let $f=z^2=2y-x^2$and also $\mathrm{Ï•}\left(x,y\right)=2x+4y-5$.

Using the Lagrange multiplier $F\left(x,y\right)=2y-x^2+\mathrm{λ}\left(2x+4y-5\right)$is to be maximised.

Differentiate $\mathrm{F}(x,y)$with respect to y and equate to 0 and solve.

$\frac{∂F\left(x,y\right)}{∂y}=2+4\mathrm{λ}$

$$\begin{gathered} 2+4\mathrm{λ}=0 \\ \mathrm{λ}=-\frac{1}{2} \end{gathered}$$

Differentiate $F\left(x,y\right)$with respect to x and equate to 0 and solve using $\mathrm{λ}=-\frac{1}{2}$.

$\frac{∂F\left(x,y\right)}{∂x}=-2x+2\mathrm{λ}$

$$\begin{gathered} -2x+2\mathrm{λ}=0 \\ -2x+2\left(-\frac{1}{2}\right)=0 \\ x=-\frac{1}{2} \end{gathered}$$

Substitute$-\frac{1}{2}$ for x into$2x+4y=5$ and solve to find the value of y.

$$\begin{gathered} 2\left(-\frac{1}{2}\right)+4y=5 \\ -1+4y=5 \\ y=\frac{3}{2} \end{gathered}$$

Substitute$-\frac{1}{2}$ for x and$\frac{3}{2}$ for y into$z^2=2y-x^2$ and solve to find the value of z.

$$\begin{gathered} z^2=2\left(\frac{3}{2}\right)-{\left(-\frac{1}{2}\right)}^2 \\ {z}^2=3+\frac{1}{4} \\ z=Â\pm \sqrt{\frac{11}{4}} \end{gathered}$$

It can be observed that the maximum value of z is $\sqrt{\frac{11}{4}}$.