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Question Number 103828 by ~blr237~ last updated on 17/Jul/20

min{ ∫_0 ^1 (x^3 −px−q)^2 dx , (p,q)∈R^2 }

min{∫01(x3−px−q)2dx,(p,q)∈R2}

Answered by bobhans last updated on 17/Jul/20

(x^3 −(px+q))^2 = x^6 −2x^3 (px+q)+(px+q)^2 = x^6 −2px^4 −2qx^3 +p^2 x^2 +2pqx+q^2 ∫_0 ^1 (x^6 −2px^4 −2qx^3 +p^2 x^2 +2pqx+q^2 ) dx = (1/7)−((2p)/5)−(q/2)+(p^2 /3)+pq+q^2 = f(p,q) (∂f/∂p) = −(2/5)+((2p)/3)+q = 0 ⇒q=(2/5)−((2p)/3) (∂f/∂q) = −(1/2)+p+2q=0⇒−(1/2)+p+(4/5)−((4p)/3)=0 (p/3) = (3/(10)) ⇒p = (9/(10)) ∧q = (2/5)−(2/3).(9/(10)) q = −(1/5).

(x3−(px+q))2=x6−2x3(px+q)+(px+q)2=x6−2px4−2qx3+p2x2+2pqx+q2∫10(x6−2px4−2qx3+p2x2+2pqx+q2)dx=17−2p5−q2+p23+pq+q2=f(p,q)∂f∂p=−25+2p3+q=0⇒q=25−2p3∂f∂q=−12+p+2q=0⇒−12+p+45−4p3=0p3=310⇒p=910∧q=25−23.910q=−15.

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