The product of two positive numbers is 363. minimize the sum of the first and three times the second.

x=33,y=11

Step-by-step explanation:

The first constraint implies x*y=363, while you are finding the minimum of x+3y. You can do this with multivariable calculus approaches, but I will assume the simplest method is desired and use single-variable calculus methods.

Since xy=363, y=363/x

Substitute this into x+3y, and you get:

f(x)=x+3(363/x)=x+1089/x

Since you have this function as a function of a single variable now, you can minimize it like any other single variable function.

f'(x)=1-1089x-2

by setting f'(x)=0, the solutions x=33 and x=-33 are found. x=-33 is an extraneous solution, since you are required to have positive numbers. So, you must investigate x=33 and confirm it is a minimum.

If x=33 is a minimum of f(x), then values less than x=33 will yield a negative value in f'(x), since the function f(x) must slope downward up until it reaches a minimum. Conversely, values greater than x=33 will yield positive values in f'(x).

1<33

f'(1)=-1088

-1088<0, so f'(1) is negative

40>33

f'(40)=511/1600

511/1600>0, so f'(40) is positive.

Since the function f(x)=1-1089/x2 has a critical point at x=33 and since values less than x=33 result in negative values of f'(x) and values greater than x=33 result in positive values of f'(x), x=33 is a local minimum of f(x).

i think it would be 43 movies that would have to be watched.

step-by-step explanation:

think about it. for the second plan, every month is 8 dollars. 8 x 12= 96. so thats the number for the second plan. now for the first plan, its annual meaning you have to pay 10 dollars every year. so now you have to get an equivalent amount of 86 since 96-10=86. 86 divided by 2 since every movie costs 2$ then you would have to watch 43 movies a year to get the same as the second plan. i really hoped this . have a wonderful day as well!

step-by-step explanation:

this is not a question