wcgc wrote:
In a certain company, the formula for maximizing profits is P = -25x^2 + 7500x,, where P is profit and x is the number of machines the company operates in its factory. What value for x will maximize P?
A) 10
B) 50
C) 150
D) 200
E) 300
This one is from IntegratedLearning.
Here's what I did:
P=-25x2+7500X => P=-50+7500X, to max. P, we need to max X. So I picked E, why isn't that correct?
The OA states: To find a maximum or minimum value of an equation with an exponent in it, you take the derivative of the equation, set it to zero, and solve. I don't really get what that means. So whoever solves it, could you plz post explanation of what the above sentence mean as well?
Here's how I did it...
\(P = -25x^2 + 7500x = -25x(x - 300)\)
For \(P\) to be positive, \(x\) must be less than 300, so option E is out.
(D): \(-25(200) (-100) = 500,000\)
(C): \(-25(150) (-150) = 562,500\)
(B): \(-25(50) (-250) = 312,500\)
(A): \(-25(10) (-290) = 72,500\)
\(P\) is maximum when \(x = 150\), so option C.
Using calculus:
\(-25x^2 + 7500x\) is maximum when \(\frac{d}{dx}(-25x^2 + 7500x) = 0\).
\(\frac{d}{dx}(-25x^2 + 7500x) = 0\)
\(-50x + 7500 = 0\)
\(50x = 7500\)
\(x = \frac{7500}{50} = 150.\)
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