Two real numbers a and b have a sum of 5. Find the maximum value of the quantity a^2b.?

Answer 1

To maximise a^2b we could also maximse the log of this expression, ln(a^2b)
This could be written as 2b ln a.
Since the two numbers add to 5, substitute b = 5-a:
2(5-a)ln a
Take the derivative and set equal to 0, or, if you prefer, graph
y = 2(5-x)lnx and see where it hits a maximum.

Answer 2

you can write
y = e^2b*lna and as a+b=5 ,b= 5-a so y=e^(10-2a)ln a
In any case a has to be>0
y´= e^(10-2a)lna[-2lna +(10-2a)/a)]

-2lna +10/a -2 =0 which cannot be solved by algebraicas
Ler´s
z= -2lna+10/a-2
lim z as a=>0+ =+infinity
lim z as a=>+infinity = -infinity
z´= -2/a-10/a^2 always <0 as a>0
so z is decreasing so there is only one root and the sign of z is
+++++++root------------
At 2 z(2)>0 and at 3(z(3)<0 so the root is between 2 and 3

Answer 3

Real number include negative numbers; therefore, there is no solution to your question. For example, if you say -95 ^ 100 is the maximum, I'd say -995 ^ 1000 is larger.

Answer 4

There is no maximum.
lim a→-∞ a^2b→(-∞)^2*∞=∞^3