Answer:
15, 15
Step-by-step explanation:
Let the first and second number be x and y.
Three times the second number = 3y
As sum of the first number and three times the second number is 30,
x + 3y =30
we get, x = 30 - 3y
We need to find two positive real numbers whose product is maximum.
Let P =xy = y(30-3y)=[tex]30y-3y^2[/tex]
0n differentiating P with respect to y, we get
[tex]P'=30-6y[/tex]
On putting [tex]P'=0[/tex], we get [tex]30-6y=0\Rightarrow 6y=30\Rightarrow y=5[/tex]
0n differentiating P ' with respect to y, we get
[tex]P''=-6[/tex]
At y = 5, [tex]P''=-6< 0[/tex]
So, y = 5 is a point of maxima.
At y = 5, [tex]x=30-3y=30-3(5)=30-15=15[/tex]
So, the two positive numbers are 15, 15
