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I will present you another approach.
From the condition, you have this system of 2 equations in 2 unknowns
3000A + 4000B = 36000 (1) (nutrient A)
1000A + 4000B = 20000 (2) (nutrient B)
To solve the system, subtract equation (2) from equation (1). You will get
2000A = 36000 - 20000 = 16000
A = 16000/2000 = 8.
Then from equation (2),
1000*8 + 4000B = 20000
4000B = 20000 - 1000*8 = 12000
B = 12000/4000 = 3.
ANSWER. A = 8, B = 3 satisfies to BOTH requirements to provide the necessary minimum nutrients and to minimize cost.
The minimum cost is thus 20 c/kilo * 8 kilograms + 40 c/kilo * 3 kilograms = 160 c + 120 c = 280 c = $2.80.
Solved.
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In the problem, the question is about "how many pounds", but I think that it is a mistake/typo,
and in reasonably posed problem the question must ask about kilograms.