gilamonster
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This isn't even funny how hard it is
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skeeter
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total surface area of a cylinder, $A = 2\pi r^2 + 2\pi rh$
from the sketch, $r = \dfrac{4x+2}{2}$ and $h=2x$
substitute for $r$ and $h$ and set equal to $182\pi$, solve for $x$, then determine the diameter
from the sketch, $r = \dfrac{4x+2}{2}$ and $h=2x$
substitute for $r$ and $h$ and set equal to $182\pi$, solve for $x$, then determine the diameter
HOI
- 921
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That's pretty straight forward. The diameter is 4x+ 2 so the radius is 2x+ 1. The top and bottom each have area $\pi r^2= \pi (2x+ 1)^2$. That totals $2\pi(2x+ 1)^2$. For the side, imagine cutting down the side and "unrolling" it. You get a rectangle with width 2x and length equal to the circumferce of the circles $2\pi r= \pi(4x+ 2)$, That area is $\pi(8x^2+ 4x)$ so the total area is $2\pi(2x+ 1)^2+ \pi(8x^2+4x)= \pi(8x^2+ 8x+ 1+ 8x^2+ 4x)= \pi(16x^2+ 12c+ 1)= 182\pi$.
Dividing both sides by $\pi$, $16x^2+12x+ 1= 182$. Subtracting 182 from both sides, $16x^2+ 12x- 181= 0$. Solve that quadratic equation for x.
Dividing both sides by $\pi$, $16x^2+12x+ 1= 182$. Subtracting 182 from both sides, $16x^2+ 12x- 181= 0$. Solve that quadratic equation for x.
skeeter
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$2\pi (2x+1)^2 + \pi(8x^3+4x) = \pi(8x^2+8x + {\color{red}2} +8x^2+4x)$