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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