A cereal company is exploring new cylinder-shaped containers. Two possible containers are shown in the diagram.x is 6 and r=4.5 y is 10.5 and r=3 If both containers have labels covering their lateral surfaces, which container will have less label area and by how much? Container Y by about 42 in.2. Container X by about 28 in.2. Container Y by about 9 in.2. Container X by about 85 in.2.

A cereal company is exploring new cylindershaped containers Two possible containers are shown in the diagramx is 6 and r45 y is 105 and r3 If both containers ha class=

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Answer:

Container [tex]\rm X[/tex] will have less label area than container [tex]\rm Y[/tex] by about [tex]9\; \rm in[/tex].

Step-by-step explanation:

A rectangular sheet of paper can be rolled into a cylinder. Conversely, the lateral surface of a cylinder can be unrolled into a rectangle- without changing the area of that surface.

Indeed, the width of that rectangle will be the same as the height of the cylinder. On the other hand, the length of that rectangle should be exactly equal to the circumference of the circles on the top and the bottom of the cylinder. In other words, if a cylinder has a height of [tex]h[/tex] and a radius of [tex]r[/tex] at the top and the bottom, then its lateral surface can be unrolled into a rectangle of width [tex]h[/tex] and length [tex]2\,\pi\, r[/tex].

Apply this reasoning to both cylinder [tex]\mathrm{X}[/tex] and [tex]\rm Y[/tex]:

For cylinder [tex]\mathrm{X}[/tex], [tex]h = 6\; \rm in[/tex] while [tex]r = 4.5\; \rm in[/tex]. Therefore, when the lateral side of this cylinder is unrolled:

  • The width of the rectangle will be [tex]6\; \rm in[/tex], while
  • The length of the rectangle will be [tex]2 \, \pi \times 4.5\; \rm in = 9\, \pi\; \rm in[/tex].

That corresponds to a lateral surface area of [tex]54\, \pi\; \rm in^2[/tex].

For cylinder [tex]\rm Y[/tex], [tex]h = 10.5\; \rm in[/tex] while [tex]r = 3\; \rm in[/tex]. Similarly, when the lateral side of this cylinder is unrolled:

  • The width of the rectangle will be [tex]10.5\; \rm in[/tex], while
  • The length of the rectangle will be [tex]2\pi\times 3\; \rm in = 6\,\pi \; \rm in[/tex].

That corresponds to a lateral surface area of [tex]63\,\pi \; \rm in^2[/tex].

Therefore, the lateral surface area of cylinder [tex]\rm X[/tex] is smaller than that of cylinder [tex]\rm Y[/tex] by [tex]9\,\pi\; \rm in^2[/tex].