Answer :
x + y = 100 ==> x = (100 - y)
P = xy^2 = (100 - y)y^2 = 100y^2 - y^3
P ' = 200 y - 3y^2
P" = 200 - 6y
P' = 0 when 200 y - 3y^2 = 0
y(200 - 3y) = 0
so either y = 0 or 200 - 3y = 0
y = 0 or y = 66.667
and since x = 100 - y, then if y = 0, then x = 100
or if y = 66.667, then x = 33.333
by inspection, if y = 0, the product is 0 (a minimum)
so (33.333 , 66.667) is the maximum
P " (66.667) < 0, confirming this is a max
P = xy^2 = (100 - y)y^2 = 100y^2 - y^3
P ' = 200 y - 3y^2
P" = 200 - 6y
P' = 0 when 200 y - 3y^2 = 0
y(200 - 3y) = 0
so either y = 0 or 200 - 3y = 0
y = 0 or y = 66.667
and since x = 100 - y, then if y = 0, then x = 100
or if y = 66.667, then x = 33.333
by inspection, if y = 0, the product is 0 (a minimum)
so (33.333 , 66.667) is the maximum
P " (66.667) < 0, confirming this is a max
The solution to the problem is as follows:
x+y = 100
f(x) = x²y = x²(100-x) = 100x² - x³
f'(x) = 200x - x² = 0 => x = 0, 200
x = 200
y = -100
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
x+y = 100
f(x) = x²y = x²(100-x) = 100x² - x³
f'(x) = 200x - x² = 0 => x = 0, 200
x = 200
y = -100
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!