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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 }

$$\:\:\:\:\:{min}\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} −{px}−{q}\right)^{\mathrm{2}} {dx}\:,\:\:\left({p},{q}\right)\in\mathbb{R}^{\mathrm{2}} \:\right\} \\ $$

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).

$$\left({x}^{\mathrm{3}} −\left({px}+{q}\right)\right)^{\mathrm{2}} =\:{x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{3}} \left({px}+{q}\right)+\left({px}+{q}\right)^{\mathrm{2}} \\ $$$$=\:{x}^{\mathrm{6}} −\mathrm{2}{px}^{\mathrm{4}} −\mathrm{2}{qx}^{\mathrm{3}} +{p}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{pqx}+{q}^{\mathrm{2}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({x}^{\mathrm{6}} −\mathrm{2}{px}^{\mathrm{4}} −\mathrm{2}{qx}^{\mathrm{3}} +{p}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{pqx}+{q}^{\mathrm{2}} \right)\:{dx}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{2}{p}}{\mathrm{5}}−\frac{{q}}{\mathrm{2}}+\frac{{p}^{\mathrm{2}} }{\mathrm{3}}+{pq}+{q}^{\mathrm{2}} \:=\:{f}\left({p},{q}\right) \\ $$$$\frac{\partial{f}}{\partial{p}}\:=\:−\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{2}{p}}{\mathrm{3}}+{q}\:=\:\mathrm{0}\:\Rightarrow{q}=\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{2}{p}}{\mathrm{3}} \\ $$$$\frac{\partial{f}}{\partial{q}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}+{p}+\mathrm{2}{q}=\mathrm{0}\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}+{p}+\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{4}{p}}{\mathrm{3}}=\mathrm{0} \\ $$$$\frac{{p}}{\mathrm{3}}\:=\:\frac{\mathrm{3}}{\mathrm{10}}\:\Rightarrow{p}\:=\:\frac{\mathrm{9}}{\mathrm{10}}\:\wedge{q}\:=\:\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{9}}{\mathrm{10}} \\ $$$${q}\:=\:−\frac{\mathrm{1}}{\mathrm{5}}.\: \\ $$

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