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:

a+b=12 y=min(a3+b3) y=min(a3+(12a)3)  f=3a23(12a)2 f=0   3 x23 (12x)2=0  72x=432  x=6  a1=6a+b=12 \ \\ y=min(a^3+b^3) \ \\ y=min(a^3+(12-a)^3) \ \\ \ \\ f'=3a^2 -3(12-a)^2 \ \\ f'=0 \ \\ \ \\ \ \\ 3 \cdot \ x^2 -3 \cdot \ (12-x)^2=0 \ \\ \ \\ 72x=432 \ \\ \ \\ x=6 \ \\ \ \\ a_{1}=6
b1=12a1=126=6b_{1}=12-a_{1}=12-6=6
f2=a b3 f2=a (12a)3 f2(a)=d/da(a(12a)3)=4(a3)(12a)2  f2(a)=0  m1=3 m2=12 m3=12  f21=m1 (12m1)3=3 (123)3=2187 f22=m2 (12m2)3=12 (1212)3=0   a2=m1=3f_{2}=a \cdot \ b^3 \ \\ f_{2}=a \cdot \ (12-a)^3 \ \\ f_{2}'(a)=d/da(a (12 - a)^3)=-4 (a - 3) (12 - a)^2 \ \\ \ \\ f_{2}'(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
b2=12a2=123=9b_{2}=12-a_{2}=12-3=9
f3=a b2 f3=a (12a)2 f3(a)=d/da(a(12a)2)=3(a216 a+48)  f3(a)=0  3(z216z+48)=0  3(z216 z+48)=0 3z248z+144=0  a=3;b=48;c=144 D=b24ac=48243144=576 D>0  z1,2=b±D2a=48±5766 z1,2=48±246 z1,2=8±4 z1=12 z2=4   Factored form of the equation:  3(z12)(z4)=0  f31=z1 (12z1)2=12 (1212)2=0 f32=z2 (12z2)2=4 (124)2=256  a3=z2=4

Our quadratic equation calculator calculates it.

b3=12a3=124=8b_{3}=12-a_{3}=12-4=8



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