Question

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?

Answer 1

`x=(20)/(3)` units and
`y=(10)/(3)` units is required to maximize the weight.

Explanation:

Given weight function is,

`W(x,y)=13xy(20-x-2y)=260xy-13x^2y-26xy^2\hfill (1)`

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.

• First we have to find the critical points, differentiate partially (1) with respect to x and make it eual to zero we get,

`W_x(x,y)=0\implies 260y-26xy-26y^2=0\implies y^2+xy-10y=0\implies y(x+y-10)=0`

`\therefore` Either, y=0 or,
`x+y=10\hfill (2)`

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

`W_y(x,y)=0\implies 260x-13x^2-52xy=0\implies 13x(20-x-4y)=0`

`\therefore` Either, x=0 or,
`x+4y=20\hfill (3)`

Solving (2) and (3) we will get,
`x=(20)/(3), y=(10)/(3)`.

Hence critical points of (1) are (0,0) and
`((20)/(3),(10)/(3)`

• Now to find maxima we have to verify both points.

Consider,

`W_(xx)=A=-26y,W_(xy)=B=260-26x-52y,W_(yy)=C=-52x`

So that,

`AC-B^2=1352-(260-26x-52y)^2`

At (0,0),

`AC-B^2=-260<0`

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

At
`((20)/(3),(10)/(3))`

`AC-B^2=30131.111>0`

Which imply
`((20)/(3),(10)/(3))` is local maxima of (1).

Therefore
`x=(20)/(3)` unit and
`y=(10)/(3)` unit is required to maximize the weight.