Answer:
Part 16) The marginal revenue is $4,000
Part 17) The marginal average cost function is (70x+41)
Explanation:
Part 16) we know that
The marginal revenue function is simply the derivative of the revenue function
we have
![R(x)=5x-0.0005x^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/ee0m3r7am6flnojhh5kz83bdo1mbd8iaf0.png)
Find
![(dR(x))/(dx)](https://img.qammunity.org/2020/formulas/mathematics/college/qfmk3v2w50jfenjkp3cvksd4au60u9760y.png)
![R'(x)=5-0.001x](https://img.qammunity.org/2020/formulas/mathematics/college/rp2zi9rh4763r8n8toyb6tsr5wpvmr17k2.png)
For x=1,000 units
substitute
![R'(x)=5-0.001(1,000)](https://img.qammunity.org/2020/formulas/mathematics/college/qszyomr2h319ppe63tl59l9rg218v8yh0c.png)
![R'(x)=5-1=\$4](https://img.qammunity.org/2020/formulas/mathematics/college/9pavgmzcf6rzv1v33sxkx2cwtyadz3znsv.png)
Remember that the units is in thousands of dollars
therefore
The marginal revenue is $4,000
Part 17) we know that
The marginal average cost function is simply the derivative of the average cost function
we have
![C(x)=(5x+3)(7x+4)](https://img.qammunity.org/2020/formulas/mathematics/college/6zn646dses11rusf0m7fj8vpg7pvrqwr5i.png)
Applying the distributive property
![C(x)=35x^2+20x+21x+12](https://img.qammunity.org/2020/formulas/mathematics/college/df75jbszm3hvlhschc526k4e74simgyo3t.png)
![C(x)=35x^2+41x+12](https://img.qammunity.org/2020/formulas/mathematics/college/hcu0tni69r8883vmzz2xnzhxnurvoesa7f.png)
Find the derivative
![C'(x)=70x+41](https://img.qammunity.org/2020/formulas/mathematics/college/gey3i0i72pbuut142j72oqdmwtgb2c0duw.png)
therefore
The marginal average cost function is (70x+41)