Answer to Question #260596 in Calculus for vipul

Question #260596

Show that the maximum value of a 2 b 2 c 2 on a sphere of radius r centered at the origin of a Cartesian abc-coordinate system in (r 2/3)3 .


1
Expert's answer
2021-11-03T18:04:51-0400

Let f(a,b,c)=a2b2c2f(a,b,c)=a^2b^2c^2

Where a2+b2+c2=r2a^2+b^2+c^2=r^2

    g(a,b,c)=a2+b2+c2\implies g(a,b,c)=a^2+b^2+c^2

Use Lagrange multipliers to find the maximum value of a2b2c2a^2b^2c^2

f(a,b,c)=λg\nabla f(a,b,c)=\lambda\nabla g

    <2ab2c2,2a2bc2,2a2b2c>=λ<2a,2b,2c>\implies<2ab^2c^2,2a^2bc^2,2a^2b^2c>=\lambda<2a,2b,2c>

    ab2c2=aλ\implies ab^2c^2=a\lambda , a2bc2=bλa^2bc^2=b\lambda , a2b2c=cλa^2b^2c=c\lambda

    λ=b2c2\implies \lambda=b^2c^2 , λ=a2c2\lambda=a^2c^2 , λ=a2b2\lambda=a^2b^2

    b2c2=a2c2=a2b2    a=b=c\implies b^2c^2=a^2c^2=a^2b^2\implies a=b=c

Substitute a=b=c in a2+b2+c2=r2a^2+b^2+c^2=r^2

a=b=c=r3a=b=c=\frac{r}{\sqrt3}

(a2b2c2)max=(r23)3(a^2b^2c^2)_{max}=(\frac{r^2}{3})^3


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