3)
from x+5y=5000, we get y=(5000-x)/5
substitute into f(x) to get:
f(x)=xy=x(5000-x)/5
4)
Take derivative and equate to zero. I get x=2500,
5)
find y using equation from 3)
6)
x*y
Compute the maximum product for two positive numbers x and y with the property that the sum of the first plus five times the second is 5000.
1) Indicate the objective equation
2) Indicate the constraint equation
So the objective = f(x)=xy and the constraint = x+5y=5000
Now number 3 says rewrite the objective function in terms of the variable x only. Would it be y=x?
4) For which value of x does the objective function attain its maximum? Do I have to take the derivative?
5) What is the corresponding value of y
6) What is the vale of the maximum product of the two numbers?
5 answers
Thank You
did you get 1000-2x/5 as a derivative for x(5000-x)/5?
Yes, your derivative is correct.
The financial officer for an accounting firm allows Dh 50,000 for computer supplies in the annual budget. After 6 months, Dh 26,200 has been spent on supplies. Is this figure within 55% of the annual budget?
1 answer