Present Delivery 2.0

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The rate at which Santa delivers presents to the children of the world can be described by the function R ( t ) = ( t 10 ) 3 + π t 2 + 2 t + e 3 R(t) = -(t-10)^{3}+\pi t^{2}+\sqrt{2}t+e^{3} , where t t is measured in hours and the rate is measured in 100000 100000 presents given per hour. At t = 0 t = 0 Santa begins to deliver presents and he stops delivering presents once the rate of present giving reaches 0. There is one local maximum and one local minimum in the rate of present giving. The time exactly half way between the times when the local minimum and local maximum occur can be expressed in the form a + π b a + \frac{\pi}{b} where a a and b b are positive integers. What is a b + 12 ab + 12 ?


The answer is 42.

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1 solution

R ( t ) = ( t 10 ) 3 + π t 2 + 2 t + e 3 R(t) = -(t-10)^3 +\pi t^2 +\sqrt{2} t + e^3

R ( t ) = 3 ( t 10 ) 2 + 2 π t + 2 R'(t) = -3(t-10)^2 + 2\pi t + \sqrt{2}

The above can be written in the form of a x 2 + b x + c ax^2 + bx +c : 3 t 2 + ( 60 + 2 π ) t + ( 300 + 2 ) -3t^2 + (60 + 2\pi)t + (-300 + \sqrt2)

Solutions to the above quadratic (maximum and minimum when R'(t) = 0) : t = b + b 2 4 a c 2 a \frac{-b + \sqrt{b^2 - 4ac}}{2a} and b b 2 4 a c 2 a \frac{-b - \sqrt{b^2 - 4ac}}{2a}

Average of times: b + b 2 4 a c 2 a + b b 2 4 a c 2 a 2 \frac{\frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a}}{2}

= b 2 a = \frac{-b}{2a}

= 60 2 π 6 = \frac{-60 - 2\pi}{-6}

= 10 + π 3 = 10 + \frac{\pi}{3}

a b + 12 = 3 × 10 + 12 = 42 ab + 12 = 3 \times 10 + 12 = 42

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