Answer:
Maximum value = 32000
Step-by-step explanation:
Let the 1st number = a
the 2nd number = b
"have a sum of 60" → a + b = 60 → a = 60 - b ... [1]
"product of one number multiplied by the square of the second number" → a × b² ... [2]
substitute [1] into [2]
a × b² = (60 - b)(b²)
= 60b² - b³
To find the maximum value of 60b² - b³ → find its stationary point by using the 1st derivative = 0
[tex]d(60b^2-b^3) = 0[/tex]
[tex]2(60b^{(2-1)})-3(b^{(3-1)})=0[/tex]
[tex]120b-3b^2=0[/tex]
[tex]3b^2=120b[/tex]
[tex]b = 40[/tex]
The value will be maximum when b = 40
[tex]value=60b^2-b^3[/tex]
[tex]=60(40)^2-(40)^3[/tex]
[tex]=32000[/tex]
