Question

Find two positive numbers with a sum of 35. These two numbers must be the highest value possible when one is put the power of 4 and multiplied by the other to the power of 3

Answer 1

Using derivatives*

Answer 2

let the numbers be x and y,x>0,y>0
0<x<35,0<y<35
x+y=35
x=35-y
\(let p=x^3y^4\)
\(p=(35-y)^3y^4\)
\[\frac{ dp }{ dy }=y^4\times3(35-y)^2(-1)+4y^3(35-y)^3=y^3(35-y)^2[-3y+4(35-y)]\]
\[=y^3(35-y)^2[140-7y]\]
\[y \neq0,y \neq35\]
\[\frac{ dp }{ dy }=0,gives 140-7y=0,y=20\]

Answer 3

when y<0,slightly
\[\frac{ dp }{ dy }=(+)(+)(+)=+\]
when y>0 slightly
\[\frac{ dp }{ dy }=(+)(+)(-)=-\]
\[\frac{ dp }{ dy }~changes~sign~as~p~ passes through y=20\]
hence there is maxima at y=20
so x=35-20=15
Required numbers are 15,20

Answer 4

correction when y<20, slightly
and y> 20 ,slightly

Answer 5

Hey, sorry for not responding earlier, but I would like to thanks you. I really appreciate your help :)

Answer 6

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