
Francisco E. answered 05/26/14
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Francisco; Civil Engineering, Math., Science, Spanish, Computers.
P = −25x2 + 300x;
To look for the maximum P Is needed to take the derivative of P dP/dx which is = -50x + 300. Making dp/dx =0 we have: 0=-50x + 300 so x will be 6 and the maximum daily profit will be 900 units