A manufacturer determines that the number of drills it can sell is given by the formula − 4p 60p 05 , where is the price of the drills in dollars and D is the number of 2 + 1 − 3 = D p
drills it sells.
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?

Correct question is;

A manufacturer determines that the number of drills it can sell is given by the formula,D = -4p2 + 160p – 305 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 if drills that can be sold?

A) $20

B) D_max = 1295 drills

Step-by-step explanation:

We are given that the price for number of drills it can sell is given by the formula, D = -4p² + 160p – 305 where p is the price of the drills in dollars.

A) To find the price for selling maximum number of drills, we will find the derivative of the function D and equate to zero.

D' = -8p + 160

Equating to zero gives;

-8p + 160 = 0

8p = 160

p = 160/8

p = $20

B) maximum number of drills will occur at p = $20.

Thus;

D_max = -4(20)² + 160(20) – 305

D_max = -1600 + 3200 - 305

D_max = 1295

i dont have an answer

step-by-step explanation:

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.