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Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_0^1 (a+bx+cx^2)^2 \, dx$.

What I did is $$ \int_0^1 (a+bx+cx^2)^2 \, dx=a^2+a \left( b+\frac{2c}{3} \right) + \frac{b^2}{3} + \frac{bc}{2}+\frac{c^2}{5}. $$ Then we consider the function $F(a,b,c)=a^2+a(b+\frac{2c}{3})+\frac{b^2}{3} + \frac{bc}{2} + \frac{c^2}{5}-\lambda(a+b+c-1)$. Thus, we can consider $\nabla F=0$ with $a+b+c=1$ and try to find $a,b$ and $c$.

However, since the function and the constraint seem to be related, I am curious whether there is some other way to see the minimum without computing Lagrange multiplier. Thank you.

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  • $\begingroup$ Look relatively what? $\endgroup$
    – Alan
    Apr 2, 2017 at 2:14
  • $\begingroup$ Sorry, it is not clear, I changed my words. $\endgroup$ Apr 2, 2017 at 2:20
  • $\begingroup$ Hint: complete the squares $ a^2 + a \left( b+\frac{2c}{3} \right) + \frac{b^2}{3} + \frac{bc}{2}+\frac{c^2}{5} $. What are the values of $ x,y,z $ if $ (x-A)^2 + (y-B)^2 + (z-C)^2 $ is minimized? $\endgroup$
    – GohP.iHan
    Apr 2, 2017 at 3:50

2 Answers 2

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Considering a quadric

$$x^2+\frac{y^2}{3}+\frac{z^2}{5}+\frac{yz}{2}+\frac{2zx}{3}+xy=\lambda$$

which touches the plane $x+y+z=1$ at $(a,b,c)$.

The tangent plane will be

$$ax+\frac{by}{3}+\frac{cz}{5}+ \frac{cy+bz}{4}+\frac{az+cx}{3}+\frac{bx+ay}{2}=\lambda$$

Comparing coefficients:

$$a+\frac{b}{2}+\frac{c}{3}= \frac{a}{2}+\frac{b}{3}+\frac{c}{4}= \frac{a}{3}+\frac{b}{4}+\frac{c}{5}=\lambda$$

gives $$(a,b,c)=3\lambda(1,-8,10)$$ and $$\lambda=\frac{1}{9}$$

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Just substitute $a=1-b-c$ and minimize the function of two variables $$g(b,c)=f(1-b-c,b,c)=\int_0^1(1-b-c+bx+cx^2)^2 dx.$$

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