Answer:
(a)![D(x)=-2,500x+60,000](https://tex.z-dn.net/?f=D%28x%29%3D-2%2C500x%2B60%2C000)
(b)![R(x)=60,000x-2500x^2](https://tex.z-dn.net/?f=R%28x%29%3D60%2C000x-2500x%5E2)
(c) x=12
(d)Optimal ticket price: $12
Maximum Revenue:$360,000
Step-by-step explanation:
The stadium holds up to 50,000 spectators.
When ticket prices were set at $12, the average attendance was 30,000.
When the ticket prices were on sale for $10, the average attendance was 35,000.
(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)
Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).
![\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500](https://tex.z-dn.net/?f=%5Ctext%7BSlope%2C%20m%7D%3D%5Cdfrac%7B30000-35000%7D%7B12-10%7D%3D-2500)
Therefore, we have:
![y=-2500x+b](https://tex.z-dn.net/?f=y%3D-2500x%2Bb)
At point (12,30000)
![30000=-2500(12)+b\\b=30000+30000\\b=60000](https://tex.z-dn.net/?f=30000%3D-2500%2812%29%2Bb%5C%5Cb%3D30000%2B30000%5C%5Cb%3D60000)
Therefore:
![D(x)=-2,500x+60,000](https://tex.z-dn.net/?f=D%28x%29%3D-2%2C500x%2B60%2C000)
(b)Revenue
![R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2](https://tex.z-dn.net/?f=R%28x%29%3Dx%20%5Ccdot%20D%28x%29%20%5Cimplies%20R%28x%29%3Dx%28-2%2C500x%2B60%2C000%29%5C%5C%5C%5CR%28x%29%3D60%2C000x-2500x%5E2)
(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.
![R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12](https://tex.z-dn.net/?f=R%28x%29%3D60%2C000x-2500x%5E2%5C%5CR%27%28x%29%3D60%2C000-5%2C000x%5C%5C60%2C000-5%2C000x%3D0%5C%5C60%2C000%3D5%2C000x%5C%5Cx%3D12)
The critical value of R(x) is x=12.
(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]
Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.
![R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0](https://tex.z-dn.net/?f=R%28x%29%3D60%2C000x-2500x%5E2%5C%5CR%281%29%3D60%2C000%281%29-2500%281%29%5E2%3D%5C%2457%2C500%5C%5CR%2812%29%3D60%2C000%2812%29-2500%2812%29%5E2%3D%5C%24360%2C000%5C%5CR%2824%29%3D60%2C000%2824%29-2500%2824%29%5E2%3D%5C%240)
Therefore:
- Optimal ticket price:$12
- Maximum Revenue:$360,000