Answer to Question #114750 in Algebra for ANJU JAYACHANDRAN

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?
1
Expert's answer
2020-05-27T18:59:15-0400

Solution:

Let the number of acres of:

maize be "x"

wheat be "y"

rice be "z"

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

From the cost per acre we get the expression;

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

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

From hours of labor per acre we get the expression;

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

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

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

Substitute (iv) in (ii);

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

"2x+3y+800-4x-4y=630"

"2x+y=170" (v)

Substitute (iv) in (iii);

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

"4x+5y+1600-8x-8y=1190"

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

Solve simultaneous equations (v) and (vi);

"4x+3y=410"

"2(2x+y=170)"

"(4x+3y=410)"

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

"y=70"

Use equation (v) to get "x";

"2x+70=170"

"2x=100"

"x=50"

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

Thus; "z=200-50-70"

"z=80"

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


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