Given the following profit function for a company, #p(s)=-20s^2 +1400s -12000#, if the selling price is $35 how much will the company profit? What's the difference in profit if the selling price is $20 verses $53?

Answer 1

At selling price of #$35# , profit is #$12500#
At selling price of #$20 and $53# profits are
#p(20)= $8000# and #p(53)= $6020#

Let #s# denotes selling price , then profit function of the company
is #p(s) -20 s^2 +1400 s -12000# . When selling price is #s=$35#
then profit is #p(35)= -20*35^2 +1400*35 -12000# or
#p(35)= -20*35^2 +1400*35 -12000= $12500#
When selling price is #s=$20# , profit is
#p(20)= -20*20^2 +1400*20 -12000= $8000#
When selling price is #s=$53# , profit is
#p(53)= -20*53^2 +1400*53 -12000= $6020# [Ans]
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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