A silo is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square foot surface area is twice as much for the hemisphere as for the cylinder. Determine the dimensions to be used if the volume is fixed (100m^3) and the cost is to be minimum.plz show stepsthanks in advance-lara
Let V = volume of silo V = volume of hemisphere + volume of cylinder Volume of hemisphere = (2/3)(pi)r^3 Volume of cylinder = (pi)r^2h where r = radius of cylinder = radius of hemisphere h = height of the cylinder Since V = 100, 100 = (2/3)(pi)r^3 + pi(r^2)h and solving for h h = (100/(pi*r^2)) - (2r/3) S = Surface area of silo = surface area of hemisphere + surface area of cylinder Surface area of hemisphere = 2(pi)r^2 Surface area of cylinder = pi(r^2) + 2(pi)rh Therefore, S = 2(pi)r^2 + pi(r^2) + 2(pi)rh since h = (100/(pi*r^2)) - (2r/3) and substituting it in the above equation, S = 2(pi)r^2 + pi(r^2) + 2(pi)r[100/(pi*r^2)) - (2r/3)] Simplifying the above, S = 2(pi)r^2 + [(pi)r^2 + 200/r - 2(pi)r^2/3] S = 2(pi)r^2 + [(1/3)(pi)r^2 + 200/r] Since cost of hemisphere is twice that of the cylinder, then C = cost = 2*2(pi)r^2 + 1*[(1/3)(pi)r^2 + 200/r] C = 4(pi)r^2 + (1/3(pi)r^2 + 200/r C = (13/3)(pi)r^2 + 200/r Differentiating C with respec to r, (dC/dr) = (26/3)(pi)r - 200/r^2 and setting (dC/dr) = 0 and solving for r, (26/3)(pi)r - 200/r^2 = 0 Simplifying the above, 26(pi)r^3 = 600 Solving for r, r = 1.94 meters Now, that r is known, h can be determined by using the above derived formula of h = (100/(pi*r^2)) - (2r/3) Substituting r = 1.94. h = 7.16 meters Dimensions of the silo are: r = 1.94 m h = 7.16 m CHECK: First check: V = (2/3)(pi)(1.94)^3 + pi(1.94)^2(7.16) V = 15.3 + 84.7 = 100 m (this checks with the original condition of the problem that the volume of the silo is fixed at 100 m^3). Second check: Take the second derivative of the original function, i.e., (d^2C/d^2r) = 400/r^3 and since the second derivative is positive (greater than zero), then it confirms that the calculated dimensions for r and h will indeed yield the minimum cost.
