Derivative problem

The sum of two numbers is 12. Find these numbers if:
a) The sum of their third powers is minimal.
b) The product of one with the cube of the other is maximal.
c) Both are positive and the product of one with the other power of the other is maximal.

Correct result:

a₁ =  6
b₁ =  6
a₂ =  3
b₂ =  9
a₃ =  4
b₃ =  8

Solution:

$$\begin{gathered} a+b=12 \\ y=min(a^3+b^3) \\ y=min(a^3+(12-a{)}^3) \\ {f}^{\mathrm{\prime }}=3a^2-3(12-a{)}^2 \\ {f}^{\mathrm{\prime }}=0 \\ 3\cdot x^2-3\cdot (12-x{)}^2=0 \\ 72x=432 \\ x=6 \\ {a}_1=6 \end{gathered}$$

$b_1=12-a_1=12-6=6$

$$\begin{gathered} f_2=a\cdot b^3 \\ {f}_2=a\cdot (12-a{)}^3 \\ {f}_2^{\mathrm{\prime }}(a)=d\mathrm{/}da(a(12-a{)}^3)=-4(a-3)(12-a{)}^2 \\ {f}_2^{\mathrm{\prime }}(a)=0 \\ {m}_1=3 \\ {m}_2=12 \\ {m}_3=12 \\ {f}_{21}=m_1\cdot (12-m_1{)}^3=3\cdot (12-3{)}^3=2187 \\ {f}_{22}=m_2\cdot (12-m_2{)}^3=12\cdot (12-12{)}^3=0 \\ {a}_2=m_1=3 \end{gathered}$$

$b_2=12-a_2=12-3=9$

$$\begin{gathered} f_3=a\cdot b^2 \\ {f}_3=a\cdot (12-a{)}^2 \\ {f}_3^{\mathrm{\prime }}(a)=d\mathrm{/}da(a(12-a{)}^2)=3(a^2-16a+48) \\ {f}_3^{\mathrm{\prime }}(a)=0 \\ 3(z^2-16z+48)=0 \\ 3(z^2-16z+48)=0 \\ 3z^2-48z+144=0 \\ a=3;b=-48;c=144 \\ D=b^2-4ac=48^2-4\cdot 3\cdot 144=576 \\ D>0 \\ {z}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{48\pm \sqrt{576}}{6}} \\ {z}_{1,2}={\displaystyle \frac{48\pm 24}{6}} \\ {z}_{1,2}=8\pm 4 \\ {z}_1=12 \\ {z}_2=4 \\ \text{ Factored form of the equation: } \\ 3(z-12)(z-4)=0 \\ {f}_{31}=z_1\cdot (12-z_1{)}^2=12\cdot (12-12{)}^2=0 \\ {f}_{32}=z_2\cdot (12-z_2{)}^2=4\cdot (12-4{)}^2=256 \\ {a}_3=z_2=4 \end{gathered}$$

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$b_3=12-a_3=12-4=8$

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