Answer to Question #102765 in Algebra for bwjsvs

Question

A farmer has 200 acres of land suitable for growing rice, wheat and maize. The cost of growing maize, wheat and rice is Rs.40/-, Rs.60/- and Rs.80/- per acre, respectively.
The farmer has Rs.12,600/- available. To grow maize she requires 20 hours of labour per acre, for wheat 25 hours of labour per acre and for rice 40 hours of labour per acre.
She has a maximum of 5950 hours of labour available. If she wants to use all her cultivable land, all the budget and all the labour available, how many acres of each crop should she plant?

Answer 1

As per the given question,

Total land =200 acres

The cost of the growing maize=40 per acre

The cost of the growing wheat=60 per acre

And cost of the growing rice =80 per acre

The total amount farmer have =12600

Maize requires total =20 hours of labour per acre

Wheat requires total =25 hours of labour per acre

Rice requires total =40 hours of labour per acre

Total hours of labour have =5950

Let the number of acres for the Maize=x

Number of acres for the crop wheat=y and

The number of acres for the rice=z

Hence, total cost

$40x+60y+80z=12600$

$4x+6y+8z=1260--------(i)$

Total labour hours

$20x+25y+40z=5950$

$4x+5y+8z=1190------(ii)$

And total available land

$x+y+z=200-------(iii)$

Now, equation (ii) is subtracting from the equation (i)

$y=70$

Now putting the value of y in the equation (ii) and in equation (iii)

$4x+5\times 70+8z=1190$

$4x+8z=1190-350=840$

$4x+8z=840-----(iv)$

$x+70+z=200$

$x+z=200-70=130$

$x+z=130$

now multiplying by 4 in the both sides,

$4x+4z=520-------(v)$

Now, subtracting equation (v) from the equation (iv)

$4z=840-520=320$

$z=\dfrac{320}{4}=80$

Now, putting the values of y and z in the equation (v)

$x=200-70-80=50$

So,

$x=50$ acres

$y=70$ acres

$z=80$ acres

So, she should plant 50acres of maize, 70 acres of wheat and 80 ares of rice.