Acompany manufactures and sells x meras per month the monthly revnuee function is given by r(x)=200x-0.02x+1000 a) what is the maximum monthly revenue? b) how many cameras should be produced to maximize the monthl revenue?

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204