Find two positive numbers satisfying the given requirements. the sum of the first and twice the second is 160 and the product is a maximum.

Two positive numbers x, y, such that x + 2y = 160.

The values of x and y that maximize xy.

The product will be maximized when its derivative is zero. We can write the product in terms of a single variable to make solution easier.

... y = (160 -x)/2 . . . . . solve the given relation for y

... f(x) = x·y = x(160 -x)/2 . . . . . substitute the expression for y into the product

... f'(x) = 80 -x = 0 . . . . . . . . . . find the derivative of f(x) and set it to zero

... x = 80 . . . . . . . . . . . . . . . . . add 80

... y = (160 -80)/2 = 40 . . . . . . find the value of y

The first number is 80.

The second number is 40.

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: