Advanced Functions Question

[jessacourt]

I've been stuck on this all night and was wondering if anyone could help?

In Canada, hundreds of thousands of cubic metres of wood are harvested each year. The function

f(x) = 1135x^4 - 8197x^3 + 15 868x^2 - 2157x + 176 608, 0 <= x <= 4, models the volume harvested, in cubic metres, from 1993 to 1997. Estimate the intervals (in years and months) when less than 185 000 m^3 were harvested.


Thank you so much in advance!

 

[stapel]

In Canada, hundreds of thousands of cubic metres of wood are harvested each year. The function f(x) = 1135x^4 - 8197x^3 + 15 868x^2 - 2157x + 176 608, 0 <= x <= 4, models the volume harvested, in cubic metres, from 1993 to 1997. Estimate the intervals (in years and months) when less than 185 000 m^3 were harvested.

When was the exact amount, f = 185000, harvested? Based on the shape of the graph of the function, when was less harvested?

When you reply, please include the recent topics of study. (This looks like a simple "plug into the graphing calculator" high-school pre-calculus exercise, but you've posted this to post-bachelor's-degree "Advanced Math", so clearly I'm confused about what's going on!)

Thank you!

 

[Ishuda]

I've been stuck on this all night and was wondering if anyone could help?

In Canada, hundreds of thousands of cubic metres of wood are harvested each year. The function

f(x) = 1135x^4 - 8197x^3 + 15 868x^2 - 2157x + 176 608, 0 <= x <= 4, models the volume harvested, in cubic metres, from 1993 to 1997. Estimate the intervals (in years and months) when less than 185 000 m^3 were harvested.


Thank you so much in advance!

I would assume that the function f(x) is the rate at which wood is being harvested. That is, the volume being harvested at the beginning of 1993 (x=0) is at the rate of 176 608 m³/yr. That being the case, the total volume harvested is the integral of f. If we let F indicate the integral of f, then the amount harvested over an interval \(\displaystyle \delta\)x (in years) is F(x+\(\displaystyle \delta\)x) minus the amount harvested at time x, F(x) or
\(\displaystyle {F(x+\delta x) - F(x)} = \delta x \frac{{F(x+\delta x) - F(x)}}{\delta x} \lt 185000\)

What does that fraction, \(\displaystyle \frac{{F(x+\delta x) - F(x)}}{\delta x}\), look like?

Since the rate varied by no more than about 5% during that period (from a plot of the f and the end point values) and was roughly 182 000 m³ of wood or, roughly, in any 1yr and 0.2 month period, the amount harvested would be less that 185000 m³. Answers more accurate than that would need to be worked out by the student.

 

[jessacourt]

When was the exact amount, f = 185000, harvested? Based on the shape of the graph of the function, when was less harvested?

When you reply, please include the recent topics of study. (This looks like a simple "plug into the graphing calculator" high-school pre-calculus exercise, but you've posted this to post-bachelor's-degree "Advanced Math", so clearly I'm confused about what's going on!)

Thank you!

Sorry about that. I was just as confused because I couldn't find the category to post. I didn't know that my grade 12 advanced functions course was really a pre-calculus course. We don't have a course titled pre-calculus at my school so I saw the "advanced" and took a chance. Now I know better and if ever I have another question, I will post it in the pre-calculus section. Also, we're not allowed to use calculators.

 

[jessacourt]

I would assume that the function f(x) is the rate at which wood is being harvested. That is, the volume being harvested at the beginning of 1993 (x=0) is at the rate of 176 608 m³/yr. That being the case, the total volume harvested is the integral of f. If we let F indicate the integral of f, then the amount harvested over an interval \(\displaystyle \delta\)x (in years) is F(x+\(\displaystyle \delta\)x) minus the amount harvested at time x, F(x) or
\(\displaystyle {F(x+\delta x) - F(x)} = \delta x \frac{{F(x+\delta x) - F(x)}}{\delta x} \lt 185000\)

What does that fraction, \(\displaystyle \frac{{F(x+\delta x) - F(x)}}{\delta x}\), look like?

Since the rate varied by no more than about 5% during that period (from a plot of the f and the end point values) and was roughly 182 000 m³ of wood or, roughly, in any 1yr and 0.2 month period, the amount harvested would be less that 185000 m³. Answers more accurate than that would need to be worked out by the student.

Thank you so much for taking the time to figure out my question. I totally understand what's going on now. Have a great day!