Minima Problema

Calculus Level 5

p p and q q are constants for which

f ( p , q ) = ∫ 0 π ( sin ⁡ x − ( p x 2 + q x ) ) 2 d x f(p,q)=\int_0^\pi \left(\sin x-(px^2+qx)\right)^2\;dx

has a minimum value. If

p + q = a π 2 + b π 3 + c π 4 + d π 5 , p+q=\frac{a}{\pi^2}+\frac{b}{\pi^3}+\frac{c}{\pi^4}+\frac{d}{\pi^5},

then what is the value of a + b + c + d a+b+c+d ?


The answer is -72.

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Tunk-Fey Ariawan by Tunk-Fey Ariawan , 35, Indonesia

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1 solution

The integral is plain easy, evaluating directly we get π / 2 + p 2 π 5 / 5 + q 2 π 3 / 3 + p q π 4 / 2 − 2 p ( π 2 − 4 ) − 2 q π \pi/2+p^{2}\pi^{5}/5+q^{2}\pi^{3}/3+pq\pi^{4}/2-2p(\pi^{2}-4)-2q\pi

Now, we differentiate with respect to p and q, we get

∂ f ∂ p = 2 p π 5 / 5 + q π 4 / 2 − 2 ( π 2 − 4 ) \dfrac{\partial f}{\partial p}=2p\pi^{5}/5+q\pi^{4}/2-2(\pi^{2}-4)

∂ f ∂ q = 2 q π 3 / 3 + p π 4 / 2 − 2 π \dfrac{\partial f}{\partial q}=2q\pi^{3}/3+p\pi^{4}/2-2\pi

Setting these to 0 and solve for p and q, we get, with some algebraic work

p + q = − 12 π 2 + 20 π 3 + 240 π 4 + − 320 π 5 p+q=\dfrac{-12}{\pi^{2}}+\dfrac{20}{\pi^{3}} +\dfrac{240}{\pi^{4}}+\dfrac{-320}{\pi^{5}}

So, − 12 + 20 + 240 − 320 = − 72 -12+20+240-320=\boxed{-72}

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