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4.6 surface area of prisms and cylinders as a marketing design speciali…

Question

4.6 surface area of prisms and cylinders
as a marketing design specialist, you are sketching a new label for a clients line of canned vegetables. the standard can that the client uses is 3 1/2 inches in diameter and 5 1/4 inches tall. what is the approximate area of the label that will wrap around the can? use 3.14 for π and round your answer to the nearest hundredth. only find the area of the label.
c = 2πr
l = 2πr
label area = in²

Explanation:

Step1: Find the radius.

The diameter $d = 3\frac{1}{2}=\frac{7}{2}$ inches. The radius $r=\frac{d}{2}=\frac{7}{4}$ inches.

Step2: Recall the formula for the lateral - surface area of a cylinder.

The area of the label that wraps around the can is the lateral - surface area of the cylinder, which is given by $A = 2\pi rh$, where $h$ is the height of the cylinder. Here, $h = 5\frac{1}{4}=\frac{21}{4}$ inches and $\pi = 3.14$.

Step3: Substitute the values into the formula.

$A=2\times3.14\times\frac{7}{4}\times\frac{21}{4}$
$A = 6.28\times\frac{7}{4}\times\frac{21}{4}$
$A=\frac{6.28\times7\times21}{16}$
$A=\frac{6.28\times147}{16}$
$A=\frac{923.16}{16}$
$A = 57.6975\approx57.70$

Answer:

$57.70$