Maximum Surface Area for Total Edge LengthDate: 07/14/2002 at 04:31:16 From: Peter Koloszar Subject: Calculus-max. and min. Dear Sir, Please help me solve the following problem: A piece of wire of total length L units is used to form the nine edges of a prism whose ends are equilateral triangles and whose other faces are rectangles. Show that the total surface area of the prism cannot exceed L^2/[2(12 - sqrt(3))] square units. Thank you for your help. Peter Koloszar. Date: 07/16/2002 at 02:57:46 From: Doctor Jeremiah Subject: Re: Calculus-max. and min. Hi Peter, The edges of the prism add up add to L because that's how long the wire is. If the prism has edges at the ends of length "a" units (because they are equilateral) and edges along the prism of length "b" units then the prism would look like: + / \ +-------+ | | | | | | | | | | | b | | | | + | | / \ | +---a---+ The total length of the edges is: L = 6a + 3b And the surface area is: S = 3(rectangle_area) + 2(triangle_area) = 3ab + a sqrt(3a^2/4) So now we have two equations with a and b as unknowns. We can get rid of one unknown by substitution: L = 6a + 3b b = L/3 - 2a S = 3ab + a sqrt(3a^2/4) <=== b = L/3 - 2a And then we can simplify the equation for surface area: S = 3ab + a sqrt(3a^2/4) <=== b = L/3 - 2a S = 3a(L/3 - 2a) + a sqrt(3a^2/4) S = aL - 6a^2 + a^2 sqrt(3/4) S = aL + a^2[sqrt(3/4) - 6] S = aL + 2 a^2[sqrt(3/4) - 6]/2 S = aL + a^2[2 sqrt(3/4) - 12]/2 S = aL + a^2[sqrt(3) - 12]/2 Now we have an equation for surface area. But we don't want that. We want the MAXIMUM surface area. To find the maximum we could graph the equation and look at the graph. The maximum is where the slope of the graph changes from positive to negative. To do that it must go through zero. When the slope is zero that is the maximum surface area. So we need to find the slope. We can do that with a derivative. S = aL + a^2[sqrt(3) - 12]/2 dS/da = L + 2a[sqrt(3) - 12]/2 dS/da = L + a[sqrt(3) - 12] Now, we can set the slope to 0 and find an equation that represents the point of maximum surface area dS/da = L + a[sqrt(3) - 12] 0 = L + a[sqrt(3) - 12] a[sqrt(3) - 12] = -L a = -L/[sqrt(3) - 12] a = L/[12 - sqrt(3)] This value of a is the value at the maximum surface area. If we put that into our equation for surface area we get the value of the maximum surface area: S = aL + a^2[sqrt(3) - 12]/2 <=== a = L/[12 - sqrt(3)] Substitute that and solve for S. It will equal the maximum value you were given. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 07/16/2002 at 04:48:22 From: Peter Koloszar Subject: Thank you (Calculus-max. and min.) Thanks a million for your extensive and prompt reply. I am really pleased and you guys are wonderful. Thank you very much. Peter Koloszar. |
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