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[tex]\mathbb{SOLUTION:}[/tex]
To simplify our calculations, let's consider 1 pig only.
Let:
P = profit ($, per pig)
S = sales ($, per pig)
p = price per pound ($/lb, per pig)
w = weight (lb)
C = cost per day to keep 1 pig ($)
t = number of days (days)
[tex]\\[/tex]
The profit is equal to the total sales minus the production cost.
[tex]\mathsf{P=S-C} \longleftarrow \textsf{working equation}[/tex]
[tex]\\[/tex]
From: Each weighing 300 pounds and the pigs gain weight at 10 pounds a day
[tex]\mathsf{w=300+10t}[/tex]
[tex]\\[/tex]
From: They sell today for $.75 a pound, but the price is falling
by $.01 a day
[tex]\mathsf{p=0.75-0.01t}[/tex]
[tex]\\[/tex]
The sales is equal to the price per pound multiplied by the weight.
[tex]\mathsf{S=pw}[/tex]
[tex]\mathsf{S=(0.75-0.01t)(300+10t)}[/tex]
[tex]\mathsf{S=225+7.5t-3t-0.1t^2}[/tex]
[tex]\mathsf{S=225+4.5t-0.1t^2}[/tex]
(Note: The unit of p is $/lb and the unit of w is lb, if we multiply p and w the resulting unit is $, which is the unit of the sales, S)
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From: It costs $.50 a day to keep one pig
[tex]\mathsf{C=0.50t}[/tex]
[tex]\\[/tex]
Substitute S and C in our working equation
[tex]\mathsf{P=S-C}[/tex]
[tex]\mathsf{P=225+4.5t-0.1t^2-0.5t}[/tex]
[tex]\mathsf{P=225+4t-0.1t^2}[/tex]
[tex]\\[/tex]
To maximize the profit, differentiate it then equate to zero.
[tex]\mathsf{\dfrac{d}{dt}(P=225+4t-0.1t^2)}[/tex]
[tex]\mathsf{\dfrac{dP}{dt}=4-0.2t}[/tex]
[tex]\mathsf{4-0.2t=0}[/tex]
[tex]\mathsf{0.2t=4}[/tex]
[tex]\mathsf{t=20}[/tex]
[tex]\\[/tex]
[tex]\mathbb{ANSWER:}[/tex]
To maximize the profit of the farmer, he should wait
[tex]\boxed{\mathsf{t=20 \ days}}[/tex]