Question

A fruit grower has $150$ acres of land available to raise two crops, $A$ and $B$. It takes one day to trim an acre of crop $A$ and two days to trim an acre of crop $B$ and there are $240$ days per year available for trimming. It takes $0.3$ day to pick an acre of crop $A$ and $0.1$ day to pick an acre of crop $B$ and there are $30$ days per year available for picking. The number of acres of each fruit that should be planted to maximize profit, assuming that the profit is $140$ per acre for crop $A$ and $235$ per acre for crop $B,$ is

• A. $29820$
• B. $35250$
• C. $29550$
• D. $25640$

Answer 1

The correct option is C $29550$

let $x$ = number of acres of crop $A$.
let $y$ = number of acres of crop $B$.
the equation we want to maximize is the profit equation.
the profit on crop $A$ is $140$ per acre so the total profit on crop $A$ would be:
$140x$
the profit on crop $B$ is $235$ per acre so the total profit on crop $B$ would be:
$235y$
So, objective function is : $z=140x+235y$
it takes $1$ day to trim an acre of crop $A$ and $2$ days to trim an acre of crop $B$ with a total of $240$ days available for trimming.
$1\times x+2\times y\le 240$
Also, $0.3$ day to pick an acre of crop $A$ and $0.1$ day to pick an acre of crop $B$ with a total of $30$ days available for picking.
$3x+y\le 300$
total number of acres has to be less than or equal to $150$.
$x+y\le 150$
So, constraints are:
$x+y\le 150$, $3x+y\le 300$, $x+2y\le 240,x,y\ge 0$
Now, As we have values of $z$

Corner Points	$z=140x+235y$
$(60,90)$	$29550$(maximum)
$(75,75)$	$28125$
$(0,120)$	$28200$
$(100,0)$	$14000$