Question #114750

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?

Expert's answer

Solution:

Let the number of acres of:

maize be xx

wheat be yy

rice be zz

Therefore; x+y+z=200x+y+z=200 (i)

From the cost per acre we get the expression;

40x+60y+80z=1260040x+60y+80z=12600

2x+3y+4z=6302x+3y+4z=630 (ii)

From hours of labor per acre we get the expression;

20x+25y+40z=595020x+25y+40z=5950

4x+5y+8z=11904x+5y+8z=1190 (iii)

Rearranging (a) we get; z=200xyz=200-x-y (iv)

Substitute (iv) in (ii);

2x+3y+4(200xy)=6302x+3y+4(200-x-y)=630

2x+3y+8004x4y=6302x+3y+800-4x-4y=630

2x+y=1702x+y=170 (v)

Substitute (iv) in (iii);

4x+5y+8(200xy)=11904x+5y+8(200-x-y)=1190

4x+5y+16008x8y=11904x+5y+1600-8x-8y=1190

4x+3y=4104x+3y=410 (vi)

Solve simultaneous equations (v) and (vi);

4x+3y=4104x+3y=410

2(2x+y=170)2(2x+y=170)

(4x+3y=410)(4x+3y=410)

(4x+2y=340)(4x+2y=340)-

y=70y=70

Use equation (v) to get xx;

2x+70=1702x+70=170

2x=1002x=100

x=50x=50

We know that z=200xyz=200-x-y

Thus; z=2005070z=200-50-70

z=80z=80

Answer: She should grow 50 acres of maize, 70 acres of wheat, and 80 acres of rice.


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