Your approach actually works, you might have made an inversion in the sign or something.
P(x)=-x²+40x-100
P'(x)=-2x+40=0 => x= 20
P(20)=-20²+40*20-100=300 QED
A manufacturer estimates that the profit P from producing x units of a commodity is P=-x^2+40x-100 dollars per week. What is the maximium profit he can realize in one week?
I was going to take the derivative of the equation and set it equal to 0 and then plug it back in but it didn't work. The answer is 300 but I do not know how to get this.
7 answers
oh ok I did I got x=-20 instead of 20. Thank you!!
but I still do not see how you got 300.
-20(2)=400 + 40(20)=800 - 100 = 1100?
-20(2)=400 + 40(20)=800 - 100 = 1100?
Vertex formula= -b/2a
= -40/-2 = 20
P(-20)=-(-20^2)+(40*-20)-100
P(-20)=-(400)+(800)-100
P(-20)=300
= -40/-2 = 20
P(-20)=-(-20^2)+(40*-20)-100
P(-20)=-(400)+(800)-100
P(-20)=300
Messed up my signs alittle bit should read
P(20)=-(20^2)+(40*20)-100
P(20)=-(400)+(800)-100
P(20)=300
P(20)=-(20^2)+(40*20)-100
P(20)=-(400)+(800)-100
P(20)=300
Exactly...
P(20)=-(400)+(800)-100 =300
P(20)=-(400)+(800)-100 =300
oh ok so its -400. Thanks I see now