A farmer is growing winter wheat - Leaving Cert Mathematics - Question 9 - 2014

To maximize the amount of wheat, we first find the derivative of the wheat equation:

$\frac{dY}{dN} = 0$

Calculating the derivative:

$\frac{dY}{dN} = 7000 + 32 - 0.2N$

Setting this equal to zero to find critical points:

$0 = 32 - 0.2N \\ N = 160 ext{ kg}$

Substituting N = 160 back into our original wheat equation:

$Y(160) = 7000 + 32(160) - 0.1(160)^2$

Calculating:

$Y(160) = 7000 + 5120 - 256$ $Y(160) = 9200 + 5120 - 256 = 9560 ext{ kg per hectare}$

First, calculate the revenue:

The revenue from selling wheat is:

$\text{Revenue} = 160 \times 9.56 ext{ (in tonnes)} = 1,529.60 ext{ euros}$

The revenue from leftover straw is:

$\text{Straw Revenue} = 75 ext{ euros}$

Total revenue:

$\text{Total Revenue} = 1,529.60 + 75 = 1,604.60 ext{ euros}$

Total costs are €1300, therefore the profit:

$\text{Profit} = 1,604.60 - 1,300 = 304.60 ext{ euros per hectare}$

The height of the marble is given by:

$h(t) = 441 - 4.9t^2$

To find when it hits the ground, set h(t) = 0:

$441 - 4.9t^2 = 0 \\ 4.9t^2 = 441 \\ t^2 = \frac{441}{4.9} \\ t = 9.3 ext{ seconds}$

Now, to find the speed:

$v(t) = h'(t) = -9.8t \\ v = -9.8(9.3) = -91.14 ext{ m/s}$

The speed just before hitting the ground is 29.4 m/s downwards.

To convert the speed from metres per second to kilometres per hour, multiply by 3.6:

$29.4 ext{ m/s} = 29.4 \times 3.6 = 105.84 ext{ km/h}$