Hello again, rom!
A farmer has 6000 ha available to plant with corn and wheat.
Each hectare of corn requires 9 L of fertilizer and 3/4 hr of labour to harvest.
Each hectare of wheat requires 3 L of fertilizer and 1 hr labour to harvest.
The farmer only has available at most 40,500 L of fertilizer
and at most 5250 h of labour for harvesting.
If $\displaystyle x$ represents the number of hectares planted with corn,
and $\displaystyle y$ represents the number of hectares planted with wheat,
(a) write the system of linear inequalities that describes the constraints
and graph the feasible region for the system.
The profits per hectare are £60 for corn and £40 for wheat.
(b) Write an equation for $\displaystyle P$, the total profit.
(c) How many hectares of each crop should be planted to maximise profit?
(d) What is the maximum profit?
This is a problem in Linear Programming ... no calculus required.
Part (a)
The $\displaystyle x$ units of corn need $\displaystyle 9x$ liters of fertilizer.
The $\displaystyle y$ units of wheat need $\displaystyle 3y$ liters of fertilizer.
The fertilizer is limited to 40,500 L.
. . So we have: .$\displaystyle 9x + 3y\:\leq \:40,500\quad\Rightarrow\quad\boxed{3x + y \:\leq \:13,500}$
The $\displaystyle x$ units of corn need $\displaystyle \frac{3}{4}x$ hours to harvest.
The $\displaystyle y$ units of wheat need $\displaystyle 1y$ hours to harvest.
The total harvest time is limit to 520 hours.
. . So we have: .$\displaystyle \boxed{\frac{3}{4}x + y \:\leq \:5250}$ Code:

13,500*
*
 *
 *
 *
5250 o *
:::* *
::::::o
:::::::* *
::::::::* *
oo* 
 4500 7000
The graph is the region in Quadrant 1, below the two slanted lines.
. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
(b) The profit is 60 per $\displaystyle x$ and 40 per $\displaystyle y.$
. . . Therefore: .$\displaystyle \boxed{P \:=\:60x + 40y}$
. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
(c) For maximum profit, we consider only the vertices of the shaded region.
We know three vertices: .$\displaystyle (4500,0),\;(0,5250)$, and $\displaystyle (0,0)$ (which we'll disregard).
The intersection of $\displaystyle 3x + y \:=\:13,500$ and $\displaystyle 3x + 4y \:=\:21,000$
. . is: .$\displaystyle \left(\frac{11,000}{3},\,2500\right)$ ... the fourth vertex.
$\displaystyle \begin{array}{ccc}(4500,\,0): \\ (0,\,5250): \\ \left(\frac{11,000}{3},\,2500\right):\end{array}
\begin{array}{ccc}P \:= \\ P\;= \\ P\:=\end{array} $ $\displaystyle \begin{array}{ccc} 60(4500) + 40(0) \\ 60(0) + 40(5250) \\ 60(\frac{11,000}{3}) + 40(2500) \end{array} $ $\displaystyle \begin{array}{ccc}= \\ = \\ =\end{array} $ $\displaystyle \begin{array}{ccc}270,000\\21,000\\320,000\end{arr ay}$
For maximum profit: .$\displaystyle \boxed{x\,=\,\frac{11,000}{3},\;y\,=\,2500}$
. . . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
(d) The maximum profit is: .$\displaystyle \boxed{\$320,000}$