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Is this correct if not how do I solve this ?

Is this correct if not how do I solve this ?-example-1
User Wangchi
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1 Answer

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we have

the revenue is equal to

R(X)=[1,500-3x]x-[66,500+20x]

R(x)=1,500x-3x^2-66,500-20x

R(X)=-3x^2+1,480x-66,500

using a graphing tool

the vertex is the point (246.67,116,033.33)

that means

the maximum revenue is For x=246.67

For break even

1,500x-3x^2=66,500+20x

the solutions are

x=50 and x=443

but x=443 is not in the domain

so

break even is for x=50

part e

For x=35

substitute in the revenue equation

R(X)=-3x^2+1,480x-66,500

R(35)=-3(35)^2+1,480(35)-66,500

R(35)=-18,375 -----> is negative, because is a loss

part f

For a profit of 100,000

we have

R(x)=100,000

substitute

100,000=-3x^2+1,480x-66,500

-3x^2+1480x-166500=0

the solution is For x=94.43

User Ben Max Rubinstein
by
8.2k points

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