document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=201229>Question 201229</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Landsville has just finished building a new water tank. The tank is a cylinder with a cone on top. It has a diameter of 20 ft., a total height of 21 ft., and the cylindrical part is 15 ft. tall. The city decided to paint the whole outside (excluding the base or bottom). The paint costs $18 per gallon and only takes one coat. Each gallon covers approximately 250 sq. ft. How many gallons should the city buy? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #151496 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=151496\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "Consider the cone part and the cylinder part as separate pieces, calculate their areas, and then sum them.\r<BR>\n" );
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document.write( "First, let's look at the cylinder.  Since the problem says to exclude the bottom of the tank (I suspect it is built on the ground), and since the cone sits on top of the cylinder so you can exclude the other (top) base of the cylinder, you are looking for the lateral surface area (defined as the surface area excluding both bases).\r<BR>\n" );
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document.write( "So let's look at the situation intuitively.  If you cut the open-ended cylinder along its height, and then unrolled it to lay flat, you would have a rectangle that had dimensions of the height of the cylinder by the circumference of the circular base.  Hence, the formula for lateral surface area of a right circular cylinder is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ L.A._{\ \ \ \ cyl}\ =\ 2\pi rh = \pi dh\">\r<BR>\n" );
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document.write( "where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large r\"> is the radius of the base (or <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large d\"> is the diameter]) and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large h\"> is the height of the cylinder.\r<BR>\n" );
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document.write( "For the cone part, you again need to consider the lateral area, that is, just the slanted surface of the cone excluding the base.  The intuitive concept is a little more challenging here, and it actually takes some Calculus to prove.  The basic idea is this:  If you draw a bunch of lines from the apex of the cone to the circumference of the base and space them evenly, you will have divided the surface area into a bunch of very skinny isosceles triangles -- or very nearly so, since the base of the triangles will not be exactly straight lines because of the curvature of the circular base.  However, if you divide it up into enough pieces, the area of each piece will be very close to the area of a perfect triangle.  Now, the sum of all of those triangles is the area we are looking for.  The area of one of the triangles is, as always, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{1}{2}bh\"> where the base is the circumference of the base of the cylinder divided by the number of skinny triangles we drew (or very very close), and the height is very nearly the same as the length of one of the sides of the triangle -- and the skinnier we make the triangles, the closer we can get.\r<BR>\n" );
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document.write( "So, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large 2\pi r\"> being the circumference of the base of the cone, the base of one of our little triangles is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{2\pi r}{n}\"> if we have <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large n\"> triangles, and the height of one of the triangles is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large s\">, the slant height, or measure from the apex to the circular base.  And since we have <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large n\"> triangles, the total lateral area is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ n\left(\frac{1}{2}\right)\left(\frac{2\pi r}{n}\right)s =  \frac{1}{2}\cdot\2\pi rs = \pi rs\">\r<BR>\n" );
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document.write( "Now the intuitive explanation above may have given you the impression that this formula only gives you an approximate answer.  That's where the Calculus would come in and prove that it is exact.  You can take that on faith, or write back and I'll give you the proof.\r<BR>\n" );
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document.write( "Now the only thing that we have left to do is figure out what the slant height, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large s\">, measures in terms of the information we are given, namely the radius and height of the cone.\r<BR>\n" );
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document.write( "If you slice the cone along a diameter, the flat surface you would see would be an isosceles triangle where the two equal sides are slant heights and the base is the diameter of the cone base.  That triangle is divided into two congruent right triangles where one leg is the height of the cone and the other is the radius of the base.  Therefore, the measure of the slant height in terms of the radius of the base and the height of the cone is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ s = \sqrt{r^2 + h^2}\">\r<BR>\n" );
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document.write( "Thank you, Mr. Pythagoras.\r<BR>\n" );
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document.write( "Now we can construct the complete formula for the lateral surface area of the cone:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ L.A._{\ \ \ \ \ \ cone}\ + =\ \pi r \sqrt{r^2 + h^2}\">\r<BR>\n" );
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document.write( "Finally, we can put the pieces together and create a formula for the total surface area of the tank:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ L.A._{\ \ \ \ \ \ cone}\ + L.A._{\ \ \ \ cyl}\ =\ \pi r \sqrt{r^2 + h_{cone}^2} + 2\pi rh_{cyl}\">\r<BR>\n" );
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document.write( "Next, let's look at the information we were given.  The overall height of the tank is 21 feet, and the height of the cylinder part is 15 feet, leaving 6 feet for the height of the cone part.  The diameter of the tank is 20 feet meaning the radius is 10 feet.  Now we can plug in the values.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ L.A._{\ \ \ \ \ \ cone}\ + L.A._{\ \ \ \ cyl}\ =\ \pi (10) \sqrt{(10)^2 + (6)^2} + 2\pi (10)(15)\">.\r<BR>\n" );
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document.write( "You can do your own arithmetic.  3.14 is certainly close enough as a value for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pi\">.  Having calculated the total lateral surface area of the tank, you will need to divide that value by 250 to get the number of gallons of paint.  Round your answer UP to the next integer -- the local Lowe's is not going to sell anyone a fractional part of a gallon of paint.\r<BR>\n" );
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document.write( "By the way, as a practical matter, the Landsville Public Works Department would probably make this calculation and then add 10% to 20% to account for spillage, overestimation of the coverage of a gallon of paint, and to have a stock of the same color to be used for maintenance touch-ups in the future.  The real world answer would probably be two 5-gallon buckets, your calculation notwithstanding.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
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