Margaret runs a business. This year’s revenue is given by the function R = −0.5x2 − 200x. Can her revenue be at least $25,000 this year?
Quadratic Math

This can be determined by using first derivative test.Therefore, her revenue  cannot be at least $25,000 this year.

Given :

Revenue  is given by the function as  R =                            ...(1)

By using of  first derivative test, Margaret revenue will be at the maximum if

By first derivative test,

                                             ...(2)

As we know that ,

Now, calculate further  from equation (2),

Now, put the value of  in equation (1) and solve it further,

Therefore,it  means that the revenue for the year is $20,000. From this we have to show that her revenue cannot be at least $25000 for that year,but she could make less than $25000.

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No, she can make less than $25000 according to the calculation

Step-by-step explanation:

Given the revenue function expressed as R = −0.5x2 − 200x, Margaret revenue will be at the maximum if dR/dx = 0

dR/dx = -2(0.5)x -200

dR/dx = -x -200

Since dR/dx = 0

-x-200 = 0

x = -200

Substitute x = -200 into the function.

R = −0.5x^2 − 200x

R = −0.5(-200)^2 − 200(-200)

R = -0.5(40000)+40000

R = -20000+40000

R = 20000

This means that the revenue for the year is $20,000 which shows that her revenue cannot be at least $25000 for that year but she could make less than $25000

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Step-by-step explanation:

if there is an odd amount of numbers, the median value is the number that is in the middle, with the same amount of numbers below and above. If there is an even amount of numbers in the list, the middle pair must be determined, added together, and divided by two to find the median value.