Yes, that looks like a reasonable method to find the range of A(n). Good job!

[Saracen Rue]

Homework Statement

Sally makes a window frame which can be modeled by the composite function $f\left(x\right)=\left[-8e^{\frac{x}{30}}+160\left\{0\le x\le a\right\},\ -8e^{\frac{-x}{30}}+160\left\{-a\le x<0\right\}\right]$, where $a$ is the x-intercept. A second composite function, $g\left(x\right)=\left[50\log _n\left(\frac{x}{2}+1\right)\left\{0\le x\le b\right\},50\log _n\left(1-\frac{x}{2}\right)\left\{-b\le x\le 0\right\}\right]$ (where $b$ is the x-coordinate of the point of intersection), models the bounds of the region which is to be filled with stained glass as shown:

Given that all units are in centimeters and that $1<n≤10$;

(a) Find the exact value of the constant, $a$

(b) Express the area of the region which is to be filled with stained glass as a function of $n$ and in terms of $b$.

(c) State the domain and range of the $A(n)$, correct to 3 decimal places

(d) The cost of stained glass is $450$ per square meter and the cost of regular glass is $200$ per square meter. If Sally can only spend $800$;

i) Find the maximum area of stained glass Sally could use correct to 4 decimal places
ii) Determine the value of $n$ for this area, correct to 1 decimal place.​

Homework Equations

Integrating between two points on a graph gives the area under the graph between said points

The Attempt at a Solution

I was able to do parts (a) and (b) without much trouble, but I'm rather stuck at part (c). I understand that the restriction on $n$ which was given as part of the question acts as the domain of $A(n)$, but what I don't understand is how exactly the range part works. The lowest possible area occurs at $n=10$ and is inclusive, which is fine. The largest area is confusing me though - I tried making $n=1$ to find the area at the point, and then specifying that that point is not inclusive (which is how I've always done this sort of question). However, when I do this I get not-defined as an answer. Can anybody help please?

 

[Mark44]

Homework Statement

Sally makes a window frame which can be modeled by the composite function $f\left(x\right)=\left[-8e^{\frac{x}{30}}+160\left\{0\le x\le a\right\},\ -8e^{\frac{-x}{30}}+160\left\{-a\le x<0\right\}\right]$, where $a$ is the x-intercept. A second composite function, $g\left(x\right)=\left[50\log _n\left(\frac{x}{2}+1\right)\left\{0\le x\le b\right\},50\log _n\left(1-\frac{x}{2}\right)\left\{-b\le x\le 0\right\}\right]$ (where $b$ is the x-coordinate of the point of intersection), models the bounds of the region which is to be filled with stained glass as shown:

Given that all units are in centimeters and that $1<n≤10$;

(a) Find the exact value of the constant, $a$

(b) Express the area of the region which is to be filled with stained glass as a function of $n$ and in terms of $b$.

(c) State the domain and range of the $A(n)$, correct to 3 decimal places

(d) The cost of stained glass is $450$ per square meter and the cost of regular glass is $200$ per square meter. If Sally can only spend $800$;

i) Find the maximum area of stained glass Sally could use correct to 4 decimal places
ii) Determine the value of $n$ for this area, correct to 1 decimal place.​

Homework Equations

Integrating between two points on a graph gives the area under the graph between said points

The Attempt at a Solution

I was able to do parts (a) and (b) without much trouble, but I'm rather stuck at part (c). I understand that the restriction on $n$ which was given as part of the question acts as the domain of $A(n)$, but what I don't understand is how exactly the range part works. The lowest possible area occurs at $n=10$ and is inclusive, which is fine. The largest area is confusing me though - I tried making $n=1$ to find the area at the point, and then specifying that that point is not inclusive (which is how I've always done this sort of question). However, when I do this I get not-defined as an answer. Can anybody help please?

What do you get for A(n)?

 

[Saracen Rue]

What do you get for A(n)?

I got $A(n)=2(∫_0^{b}(50\log _a\left(\frac{x}{2}+1\right))dx + ∫_b^{a}(-8e^{\frac{x}{30}}+160)dx)$ where $a=30ln(20)$

After integrating I got $A(n)=\frac{20}{\ln \left(n\right)}\left(24\left(\ln \left(n\right)\right)e^{\frac{b}{30}}+5b\ln \left(b+2\right)-16b\ln \left(n\right)+10\ln \left(b+2\right)-5b\ln \left(2\right)-5b+480\ln \left(n\right)\ln \left(5\right)+960\ln \left(n\right)\ln \left(2\right)-480\ln \left(n\right)-10\ln \left(2\right)\right)$

 

[Saracen Rue]

Bump

 

[Mark44]

I got $A(n)=2(∫_0^{b}(50\log _a\left(\frac{x}{2}+1\right))dx + ∫_b^{a}(-8e^{\frac{x}{30}}+160)dx)$ where $a=30ln(20)$

After integrating I got $A(n)=\frac{20}{\ln \left(n\right)}\left(24\left(\ln \left(n\right)\right)e^{\frac{b}{30}}+5b\ln \left(b+2\right)-16b\ln \left(n\right)+10\ln \left(b+2\right)-5b\ln \left(2\right)-5b+480\ln \left(n\right)\ln \left(5\right)+960\ln \left(n\right)\ln \left(2\right)-480\ln \left(n\right)-10\ln \left(2\right)\right)$

This looks reasonable, but I haven't worked the problem.

For part c, the domain of A(n) seems straightforward, but finding the range isn't. Your formula for area is complicated enough that you will probably need to find the max and min using graphing software. The instruction to get the domain and range correct to three decimal places seems to be a hint that you will need to use graphical techniques rather than analytic techniques.

Some things I would look at:
1) Is A(n) either increasing or decreasing on its domain? (Probably not, though.) If it is increasing, the maximum area would occur at the right end of the domain, and the minimum area would occur at the left end. If it's decreasing, the minimum area would occur at the opposite ends as described above.
2) Plot a graph of A(n) using a graphing calculator or graphing software, and look for high and low points. You could also do this using a spreadsheet for calculations. For the maximum value of A(n), if you narrow the interval you're looking at sufficiently, you should be able to get the maximum value with three decimal place precision. Same strategy for the minimum, although it might be that the minimum area is 0.

 

[Saracen Rue]

This looks reasonable, but I haven't worked the problem.

For part c, the domain of A(n) seems straightforward, but finding the range isn't. Your formula for area is complicated enough that you will probably need to find the max and min using graphing software. The instruction to get the domain and range correct to three decimal places seems to be a hint that you will need to use graphical techniques rather than analytic techniques.

Some things I would look at:
1) Is A(n) either increasing or decreasing on its domain? (Probably not, though.) If it is increasing, the maximum area would occur at the right end of the domain, and the minimum area would occur at the left end. If it's decreasing, the minimum area would occur at the opposite ends as described above.
2) Plot a graph of A(n) using a graphing calculator or graphing software, and look for high and low points. You could also do this using a spreadsheet for calculations. For the maximum value of A(n), if you narrow the interval you're looking at sufficiently, you should be able to get the maximum value with three decimal place precision. Same strategy for the minimum, although it might be that the minimum area is 0.

Graphing does seem like the easiest way to solve this problem, thank you. However I think I may have worked out another method.

For $A(n)=\frac{20}{\ln \left(n\right)}\left(24\left(\ln \left(n\right)\right)e^{\frac{b}{30}}+5b\ln \left(b+2\right)-16b\ln \left(n\right)+10\ln \left(b+2\right)-5b\ln \left(2\right)-5b+480\ln \left(n\right)\ln \left(5\right)+960\ln \left(n\right)\ln \left(2\right)-480\ln \left(n\right)-10\ln \left(2\right)\right)$, we can see there is a dilation factor of $\frac{20}{\ln \left(n\right)}$. As $log_m(1)=0$, we can conclude that at $n=1$ the dilation factor becomes $20/0$, which makes the entire function $A(n)$ not defined. So let's look at what happens when $n→1$ instead. As $n→1$, $ln(n)→0$ and $\frac{20}{\ln \left(n\right)}→∞$. From this we can predict that the largest possible area is increasing as $n$ approaches $1$. To confirm this, we can substitute various decreasing values of $n$ into $g(x)$ (such as $n=5, n=2, n=1.5, n=1.2$) and solve simultaneously with $f(x)$ to find $b$. By doing this we can see that as $n→1$, $b→0$. Looking back at my initial area equation, $A(n)=2(∫_0^{b}(50\log _n\left(\frac{x}{2}+1\right))dx + ∫_b^{30ln(20)}(-8e^{\frac{x}{30}}+160)dx)$, if we let $b=0$ we get $A(n)=2(∫_0^{0}(50\log _n\left(\frac{x}{2}+1\right))dx + ∫_0^{30ln(20)}(-8e^{\frac{x}{30}}+160)dx) =2(∫_0^{30ln(20)}(-8e^{\frac{x}{30}}+160)dx)$ Thus we can finally see that as $n→1$, $A(n)→∫_{-30ln(20)}^{30ln(20)}f(x)dx$.

Now if we actually answer the question...
$Domain_{A(n)}: (1, 10]$
$Range_{A(n)}: [A(10), 2∫_0^{30ln(20)}(-8e^{\frac{x}{30}}+160)dx) = [9874.198, 19639.030)$

Does this look like it could be right to you?