Amanufacturer determines that the number of drills it can sell is given by the formula d=-4p^2+160p-270, where p is the price of the drills in dollars. a. at what price will the manufacturer sell the maximum number of drills? b. what is the maximum number of drills that can be sold?

D'(p) = -8p + 160 

-8p + 160 = 0 

-8p = -160 

p = 20 

~~~ 
B) Number of drills sold when p = 20 is 

D(20) = [ -4(20)^2 + 160(20) - 270 ] = 1,130 drills 

I think this is right, i tried.

That is the vertex
if we complete the square
to get y=a(x-h)^2+k form wher (h,k) Is vertex
D=-4(p²-40p)-270
D=-4(p²-40p+400-400)-270
D=-4((p-20)²-400)-270
D=-4(p-20)²+1600-270
D=-4(p-20)²+1330
vertex is (20,1330)
that is at p=20 and D=1330

A. at the price of 20 units
b. can sell 1330 drills

or you can use the calc way
take deritivive to find where the slope equals 0

D'(x)=-8x+160
0=-8x+160
8x=160
x=20

we know this is the max because D'(15) is positive and D'(25) is negative so therefor at x=20, that is the max
to find the max number of drills, we do
D(20)=-4(20)²+160(20)-270
D(20)=1330

a. 20
b. 1330