Respuesta :
Answer:
A) Radius: 3.44 cm.
Height: 6.88 cm.
B) Radius: 2.73 cm.
Height: 10.92 cm.
Step-by-step explanation:
We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.
a) We can express the volume of the soda can as:
[tex]V=\pi r^2h=256[/tex]
This is the constraint.
The function we want to minimize is the surface, and it can be expressed as:
[tex]S=2\pi rh+2\pi r^2[/tex]
To solve this, we can express h in function of r:
[tex]V=\pi r^2h=256\\\\h=\frac{256}{\pi r^2}[/tex]
And replace it in the surface equation
[tex]S=2\pi rh+2\pi r^2=2\pi r(\frac{256}{\pi r^2})+2\pi r^2=\frac{512}{r} +2\pi r^2[/tex]
To optimize the function, we derive and equal to zero
[tex]\frac{dS}{dr}=512*(-1)*r^{-2}+4\pi r=0\\\\\frac{-512}{r^2}+4\pi r=0\\\\r^3=\frac{512}{4\pi} \\\\r=\sqrt[3]{\frac{512}{4\pi} } =\sqrt[3]{40.74 }=3.44[/tex]
The radius that minimizes the surface is r=3.44 cm.
The height is then
[tex]h=\frac{256}{\pi r^2}=\frac{256}{\pi (3.44)^2}=6.88[/tex]
The height that minimizes the surface is h=6.88 cm.
b) The new equation for the real surface is:
[tex]S=2\pi rh+2*(2\pi r^2)=2\pi rh+4\pi r^2[/tex]
We derive and equal to zero
[tex]\frac{dS}{dr}=512*(-1)*r^{-2}+8\pi r=0\\\\\frac{-512}{r^2}+8\pi r=0\\\\r^3=\frac{512}{8\pi} \\\\r=\sqrt[3]{\frac{512}{8\pi}}=\sqrt[3]{20.37}=2.73[/tex]
The radius that minimizes the real surface is r=2.73 cm.
The height is then
[tex]h=\frac{256}{\pi r^2}=\frac{256}{\pi (2.73)^2}=10.92[/tex]
The height that minimizes the real surface is h=10.92 cm.
