I am making a can with a hemispherical top. ( So like a silo and INCLUDE THE FLOOR).The total volume of this silo/can is 864cm^3 and I want to MINIMIZE the surface area.please show me the work
That's not a related rates problem. Are you sure you should be taking the calculus?
a million. enable h be the poster's top, and w be the poster's width. So the revealed area has top h-sixteen and width w-4, so its area is: (h - sixteen)(w - 4) = 380 w - 4 = 380 / (h - sixteen) w = (380 / (h - sixteen)) + 4 w = (380 + 4(h - sixteen)) / (h - sixteen) w = (380 + 4h - sixty 4) / (h - sixteen) w = (4h + 316) / (h - sixteen) the area of the full poster is: A = h * w A = h * (4h + 316) / (h - sixteen) A = (4h^2 + 316h) / (h - sixteen) Use the Quotient Rule to discover dA/dh: dA/dh = ((8h + 316)(h - sixteen) - (4h^2 + 316h)(a million)) / ((h - sixteen)^2) dA/dh = (8h^2 - 128h + 316h - 5056 - 4h^2 - 316h) / ((h - sixteen)^2) dA/dh = (4h^2 - 128h - 5056) / ((h - sixteen)^2) to shrink the area, set the spinoff to 0: 0 = (4h^2 - 128h - 5056) / ((h - sixteen)^2) 0 = 4h^2 - 128h - 5056 0 = h^2 - 32h - 1264 h = (-(-32) +/- sqrt((-32)^2 - 4(a million)(-1264))) / (2*a million) h = (32 +/- sqrt(1024 + 5056)) / 2 h = (32 +/- sqrt(6080)) / 2 h = sixteen +/- sqrt(1520) h =~ fifty 5 or -23 yet a unfavorable top would not make sense, so h =~ fifty 5 w = sqrt(ninety 5) + 4 w =~ 13.7 A = 444 + sqrt(97280) A =~ 755.897
Let r = radius of the circular base and hemisphere Let h = height of the silo (not including the hemisphere cap) Therefore vol of hemisphere = 1/2* 4/3 pi r^3 vol of rest of silo = pi r^2 h so total volume = 864 = 4/6 pi r^3 + pi r^2 h = pi r^2 (h+2/3r) h = 864/(pi r^2) - 2/3r SA of hemisphere = 1/2 * 4 pi r^2 SA of rest of silo = pi r^2 + 2 pi r h = pi r (r + 2h) so total SA = pi r (r + 2h + 2r) = pi r (3r +h) Plug in h Total SA = pi r (3r + 864/(pi r^2) - 2/3r) = 7/3 pi r^2 + 864r^-1 To find a minimum, differentiate and set it to 0 -> 14/3 r - 864 r^-2 =0 solve for r: test if r=5 14/3*5-864/(5*5)=-11.2 test if r=6 14/3*6-864/(6*6)= 4 so r is between 5 and 6 cm From a graphic calculator r is approx 5.7 cm, therefore h = 4.66 Checking Vol = pi r^2 (h+2/3r) = 863.52 SA = pi r (3r +h) = 389.65 This also makes sense as a sphere (and hemisphere) have a high volume to SA so it would be big relative to the silo height.
