Respuesta :
Answer:
a) Radius: 3.13 cm
Height: 9.77 cm
Surface: 122.82 cm2
b) Attached
c) To calculate the optimal point for our function, we use the property that an extreme point (maximum or minimum) can be calculated for the point in which the first derivative is equal to zero. This assumes both the function and its derivatives are continous in the range of analysis.
Step-by-step explanation:
The can has a volume of 300 cm3.
We have to minimize the surface of the can.
The surface (objective function, in this problem) can be expressed as:
[tex]S=2(\pi r^2)+2\pi h=2\pi(r^2+h)[/tex]
The volume (restriction, in this problem) can be written as:
[tex]V=\pi r^2h=300[/tex]
We can express one variable in funtion of the other:
[tex]\pi r^2h=300\\\\h=\frac{300}{\pi r^2}\\\\r^2=\frac{300}{\pi h}[/tex]
Then, the function S becomes:
[tex]S=2\pi(r^2+h)=2\pi(r^2+\frac{300}{\pi r^2})=2\pi(\frac{300}{\pi h}+h)[/tex]
Then, we derive and equal to zero
[tex]dS/dh=2\pi((-1)\frac{300}{\pi h^2} +1)=0\\\\\frac{300}{\pi h^2} =1\\\\ \pi h^2=300\\\\h=\sqrt{300/\pi} =9.77\\\\r=\sqrt{\frac{300}{\pi h} }=3.13[/tex]
In the attachment:
Black line: constraint (Volume=300)
Red lines: Contour lines for the Surface function (S=90, S=110, S=140)
Green line: Contour line for the optimum Surface (S=122)
![Ver imagen mtosi17](https://us-static.z-dn.net/files/dfe/f29f521b2cb3bedac6d949882c564b8a.png)