Mobistar is a mobile services company that sells 800
obile services company that sells 800 phones each week
when it charges $80 per phone. it sells 40 more phones per week for each
$2 decrease in price. the company's revenue is the product of the
number of phones sold and the price of each phone. what price should
the company charge to maximize its revenue?

The revenue of the company will be maximum when the price of each phone will be $60.

Step-by-step explanation:

The company charges $80 per phone and sells 800 phones each week .

Now, if the price of each phone is reduced by $2 then it sells 40 more phones per week.

Therefore, for $2x decrease in price the number of phones sold per week will be (800 + 40x)

Therefore, the revenue of the company will be given by the function

R(x) = (800 + 40x)(80 - 2x)

R(x) = 64000 - 1600x + 3200x - 80x²

R(x) = 64000 + 1600x - 80x² (1)

The condition for maximum revenue is R'(x) = 0

So, differentiating equation (1) with respect to x we get

R'(x) = 1600 - 160x = 0

⇒ x = 10

Therefore, the revenue of the company will be maximum when the price of each phone will be $(80 - 10 × 2) = $60 (Answer)

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕