Question

Find the largest $\mathit{z}$for which$\mathbf{2}\mathit{x}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{5}$ and ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{y}$.

Short answer

The maximum value of is $\sqrt{\frac{11}{4}}$.

Step-by-step solution

Given Information

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

Important information

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

Find the maximum z

Let $f=z^2=2y-x^2$and also $\varphi \left(x,y\right)=2x+4y-5$.

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

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

$\frac{\partial F\left(x,y\right)}{\partial y}=2+4\lambda$

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

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

$\frac{\partial F\left(x,y\right)}{\partial x}=-2x+2\lambda$

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

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.

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

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