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

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2