Simple differentiation (I think) Maximum problem? Basic calc.?

Find value of t @ which pt a is maximum? : p(t) (12*10^5)t^2(e^1000t)

Does that question make any sense? I don't understand what point a is? Is it just asking to find maximum? In which case I would just differentiate correct? Thanks.

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  • 9 years ago
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    P(t)=(12*10^5)*t²*e^1000t

    P'(t)=2(12*10^5)t*e^1000t+1000(12*10^5)t²*e^1000t

    P'(t)=[(24*10^5)t*e^1000t](1+500t)

    P'(t)=0 gives t=-1/500

    I think t in this question should be time.

    P''(t)=24*10^5*e^1000t*(5*10^5*t²+2000t+1)

    P''(t)<0 thus x=-1/500 gives a maximum

    where y=24/5e²

  • 4 years ago

    whilst utilising implicit differentiation, you may desire to derive the two variables in many circumstances, however denote dy/dx because of fact the by-manufactured from y each time y is derived. i'm going to apply y' to indicate dy/dx. cos(x-y)=y(2x+a million)^3 <-----Derive like favourite, interior function, exterior function, etc -(a million - y')sin(x-y) = y'(2x+a million)^3 + 6y(2x+a million)^2 <------- whilst deriving y, you get a million*y'; on the main wonderful element, you have a function circumstances a function; now simplify -sin(x-y) + y'sin(x-y) = y'(2x+a million)^3 + 6y(2x+a million)^2 <------ Now get the y's on the same element y'sin(x-y) - y'(2x+a million)^3 = 6y(2x+a million)^2 + sin(x-y) <------ element out the y' y'[sin(x-y) - (2x+a million)^3] = 6y(2x+a million)^2 + sin(x-y) <------- Divide y' = [6y(2x+a million)^2 + sin(x-y)]/[sin(x-y) - (2x+a million)^3] desire I helped!

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