A new food is designed to add weight to mature beef cattle. The weight in pounds is given by W = 13xy(20 − x − 2y), where x is the number of units of the first ingredient and y is the number of units of the second ingredient. How many units of each ingredient will maximize the weight?

units and units is required to maximize the weight.

Step-by-step explanation:

Given weight function is,

where x and y are number of units of first and second ingredients respectively.

To find how many units of each ingredient will maximize the weight we have to find the maximum value of weight function.

Either, y=0 or,

And, differentiate partially (1) with respect to y and make it equal to zero we get,

Either, x=0 or,

Solving (2) and (3) we will get, .

Hence critical points of (1) are (0,0) and

Consider,

So that,

At (0,0),

that is (0,0) is a local minima.

At

Which imply is local maxima of (1).

Therefore unit and unit is required to maximize the weight.

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743