The radius and height, in two decimal places are mathematically given as
r=10.60
Generally, the equation for the volume is mathematically given as
[tex]V=\pi r^{2} h+\frac{2}{3} \pi r^{3}[/tex]
Therefore
[tex]5000 &=\pi 0^{2} h+\frac{2}{3} \pi r^{3}[/tex]
[tex]V(r, h) &=\pi r^{2} h+\frac{2}{3} \pi r^{3}-5000 \\\nabla V &=\left\langle 2 \pi r h+2 \pi r^{2}, \pi r^{2}\right\rangle \\[/tex]
[tex]\cos t &=100\left(2 \pi r^{2}\right)+75(2 \pi r h) \\\\\cos t &=200 \pi r^{2}+150 \pi r h \\\\\nabla C(r, h) &=\langle 400 \pi r+150 \pi h, 150 \pi r\rangle[/tex]
Using Lagrange method
[tex]$\langle 400 \pi \gamma+150 \pi \gamma h, 150 \pi \gamma\rangle=\lambda\left\langle 2 \pi \gamma h+2 \pi \gamma^{2}, \pi r^{2}\right\rangle$[/tex]
[tex]$400 \pi r+150 \pi r=\left(2 \pi r h+2 \pi r^{2}\right) \lambda$[/tex]
[tex]$400 \pi \gamma+150 \pi h=300 \pi h+300 \pi r \\\\[/tex]
[tex]$400 \pi r-300 \pi r=150 \pi h$[/tex]
[tex]$160 \pi r=150 \pi h \quad \\\\\\[/tex]
with
[tex]h=\frac{2}{3} r $[/tex]
[tex]$\pi r^{2} h+\frac{2}{3} \pi r^{3}=5000$[/tex]
[tex]\pi r}\left(\frac{2}{3} r \right)+\frac{2}{3} \pi^{3}=5000$[/tex]
[tex]$\frac{4 \pi \gamma^{3}}{3}=5000 \quad[/tex]
[tex]r^{3}=\left(\frac{5000 \times 3}{4 \pi}\right)$[/tex]
r=10.60
In conclusion, the radius and height, in two decimal places is
r=10.60
Read more about radius
https://brainly.com/question/13449316
#SPJ1
