min-0-1-x-3-px-q-2-dx-p-q-R-2-

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

m i n {\int_{0}^{1} {(x^{3} - p x - q)}^{2} d x, (p, q) \in R^{2}}

Answered by bobhans last updated on 17/Jul/20

{(x^{3} - (p x + q))}^{2} = x^{6} - 2 x^{3} (p x + q) + {(p x + q)}^{2}

= x^{6} - 2 {p x}^{4} - 2 {q x}^{3} + p^{2} x^{2} + 2 p q x + q^{2}

\underset{0}{\int^{1}} (x^{6} - 2 {p x}^{4} - 2 {q x}^{3} + p^{2} x^{2} + 2 p q x + q^{2}) d x =

\frac{1}{7} - \frac{2 p}{5} - \frac{q}{2} + \frac{p^{2}}{3} + p q + q^{2} = f (p, q)

\frac{\partial f}{\partial p} = - \frac{2}{5} + \frac{2 p}{3} + q = 0 \Rightarrow q = \frac{2}{5} - \frac{2 p}{3}

\frac{\partial f}{\partial q} = - \frac{1}{2} + p + 2 q = 0 \Rightarrow - \frac{1}{2} + p + \frac{4}{5} - \frac{4 p}{3} = 0

\frac{p}{3} = \frac{3}{10} \Rightarrow p = \frac{9}{10} \land q = \frac{2}{5} - \frac{2}{3} . \frac{9}{10}

q = - \frac{1}{5} .

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