A few days ago
Use optimisation to calculate the price that will yeild the maximum total revenue for product2z”,?
where total revenue (r)=)=1200p-3p(2)
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A few days ago
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R(p) = 1200p – 3p^2
Take the derivative
R'(p) = 1200 – 6p
Set equal to zero.
1200 – 6p = 0
1200 = 6p
1200/6 = p
200 = p
The maximum revenue occurs when the price is 200.
The actual revenue can be found by evaluating R(p) at p=200.
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