Respuesta :
Answer by BlueSky06
The equation described above can also be written as, y = -x² + 100x + 4000To get the number of notebooks that will give them the maximum profit, we derive the equation and equate to zero. dy/dx = -2x + 100 = 0The value of x from the equation is 50. Then, we substitute 50 to the original equation to get the profit. y = -(50^2) + 100(50) + 4000 = 6500Thus, the maximum profit that the company makes is $6,500/day.
Read more on Brainly.com - https://brainly.com/question/3586459#readmore
The equation described above can also be written as, y = -x² + 100x + 4000To get the number of notebooks that will give them the maximum profit, we derive the equation and equate to zero. dy/dx = -2x + 100 = 0The value of x from the equation is 50. Then, we substitute 50 to the original equation to get the profit. y = -(50^2) + 100(50) + 4000 = 6500Thus, the maximum profit that the company makes is $6,500/day.
Read more on Brainly.com - https://brainly.com/question/3586459#readmore
Answer: $6,600
Step-by-step explanation:
Given: The profit function is given by the equation:
[tex]y=-x^2+120x+3000[/tex],where x is the number of netbooks produced in one day and y is the profit (in dollars) the company makes that day.
To find the maximum profit, first we need to find critical point of the function.
For that, first differentiate the above function w.r.t. x, we get
[tex]y'=-2x+120[/tex]
Now, Put [tex]y'=0[/tex], we get
[tex]-2x+120=0\\\\\Rightarrow\ x=60[/tex]
Again differentiate the above function w.r.t. x, we get
[tex]y"=-2<0[/tex]
By second derivative test, at x=60 , attains its maximum value.
Now, the Maximum profit will be :
[tex]y=-(60)^2+120(60)+3,000=6,600[/tex]
Hence, the maximum profit that the company can make = $6,600