Max Area of Triangle: x & y Axes & e^{-5x} Tangents

[Petrus]

considering the set of triangles, whose sides are the x and y axes, and the tangents to the curve \(\displaystyle e^{-5x}, x>0\) to estimate the maximum area of such a triangle can be.
I have no progress, well I know area is \(\displaystyle \frac{x*y}{2}\).

 

[Ackbach]

Re: Triangel max area

Can you get a handle on the tangent lines? Suppose I asked you to find the tangent line to $e^{-5x}$ at the value $x=1$. How would you get that? And how would you get it if I asked you to find the equation of the tangent line at $x=2$? How about an arbitrary $x$?

 

[Petrus]

Re: Triangel max area

Can you get a handle on the tangent lines? Suppose I asked you to find the tangent line to $e^{-5x}$ at the value $x=1$. How would you get that? And how would you get it if I asked you to find the equation of the tangent line at $x=2$? How about an arbitrary $x$?

for the first problem i put in x=1 in the orginal function to get y then i derivate it and put x=1 to get the slope, the derivate of the function will be \(\displaystyle -5e^{-5x}\) and put x =1 to get the slope. let's simpliefie and say y=0 and m(slope)=2 then we got the tangent equation \(\displaystyle y-y_1=m(x-x_1)\) if we take with those number i said and we get the tangent equation \(\displaystyle y=2x-2\)

 

[MarkFL]

Let's use \(\displaystyle f(x)=e^{-5x}\). You have correctly differentiated to find:

\(\displaystyle f'(x)=-5e^{-5x}\)

However, you have evaluated \(\displaystyle f(0)\) incorrectly. Can you see what you did wrong?

 

[Petrus]

Let's use \(\displaystyle f(x)=e^{-5x}\). You have correctly differentiated to find:

\(\displaystyle f'(x)=-5e^{-5x}\)

However, you have evaluated \(\displaystyle f(0)\) incorrectly. Can you see what you did wrong?

Yes I do. When x is zero Then the function \(\displaystyle e^{-5x}=1\) so I Will gave -5

 

[MarkFL]

Yes, correct! (Yes)

Can you use the point-slope formula to find the equation of the tangent line for an arbitrary value of $x$ as suggested by Ackbach, and then express this line in the two-intercept form? Do we need to worry about the signs of the intercepts?

 

[Petrus]

Yes, correct! (Yes)

Can you use the point-slope formula to find the equation of the tangent line for an arbitrary value of $x$ as suggested by Ackbach, and then express this line in the two-intercept form? Do we need to worry about the signs of the intercepts?

Well I could make a formula \(\displaystyle y=-5e^{-5x}(x_1-x)-e^{-5x}\) is this correct? Where x is our point

 

[soroban]

Hello, Petrus!

Consider the set of triangles whose sides are the x- and y-axes,
and a tangent to the curve \(\displaystyle y = e^{-5x}, x>0\).
Find the maximum area of these triangles.

Let [tex]P[/tex] be [tex]\left(p,\,e^{-5p}\right)[/tex] be any point on the curve.

The derivative is: .[tex]y' \,=\,-5e^{-5x}[/tex]

We have point [tex]P(p,\,e^{-5p})[/tex] and slope [tex]m = -5e^{-5p}[/tex]

The equation of the tangent at [tex]P[/tex] is: .[tex] y - e^{-5p} \:=\:-5e^{-5p}(x-p) [/tex]

. . which simplifies to: .[tex]y \;=\;-5e^{-5p}x + e^{-5p}(5p+1)[/tex]

The x-intercept is: .[tex]\frac{5p+1}{5}[/tex]

The y-intercept is: .[tex]e^{-5p}(5p+1)[/tex]

The area of the triangle is: .[tex]A \;=\;\frac{1}{2}\cdot\frac{5p+1}{5}\cdot e^{-5p}(5p+1) [/tex]
Hence: .[tex]A \;=\;\tfrac{1}{10}e^{-5p}(5p+1)^2[/tex] .[1]Set [tex]A'[/tex] equal to zero.

[tex]A' \;=\; \tfrac{1}{10}\left[e^{-5p}2(5p+1)5 - 5e^{-5p}(5p+1)^2\right] \;=\;0[/tex]

. . . [tex]\tfrac{1}{10}\cdot 5e^{-5p}(5p+1)\big[2 - (5p+1)\big] \;=\;0[/tex]

. . . . . . [tex]\tfrac{1}{2}e^{-5p}(5p+1)(1-5p) \;=\;0[/tex]

Hence: .[tex]p = \tfrac{1}{5},\; \color{red}{\rlap{//////}}p = \text{-}\tfrac{1}{5}[/tex]Substitute into [1]: .[tex]A \;=\;\tfrac{1}{10}e^{-5(\frac{1}{5})}\left(5[\tfrac{1}{5}] + 1\right)^2 \;=\; \tfrac{1}{10}e^{-1}(4)[/tex]

Therefore: .[tex]\text{max }A \:=\:\frac{2}{5e}[/tex]

 

[Prove It]

Hello, Petrus!


Let [tex]P[/tex] be [tex]\left(p,\,e^{-5p}\right)[/tex] be any point on the curve.

The derivative is: .[tex]y' \,=\,-5e^{-5x}[/tex]

We have point [tex]P(p,\,e^{-5p})[/tex] and slope [tex]m = -5e^{-5p}[/tex]

The equation of the tangent at [tex]P[/tex] is: .[tex] y - e^{-5p} \:=\:-5e^{-5p}(x-p) [/tex]

. . which simplifies to: .[tex]y \;=\;-5e^{-5p}x + e^{-5p}(5p+1)[/tex]

The x-intercept is: .[tex]\frac{5p+1}{5}[/tex]

The y-intercept is: .[tex]e^{-5p}(5p+1)[/tex]

The area of the triangle is: .[tex]A \;=\;\frac{1}{2}\cdot\frac{5p+1}{5}\cdot e^{-5p}(5p+1) [/tex]
Hence: .[tex]A \;=\;\tfrac{1}{10}e^{-5p}(5p+1)^2[/tex] .[1]Set [tex]A'[/tex] equal to zero.

[tex]A' \;=\; \tfrac{1}{10}\left[e^{-5p}2(5p+1)5 - 5e^{-5p}(5p+1)^2\right] \;=\;0[/tex]

. . . [tex]\tfrac{1}{10}\cdot 5e^{-5p}(5p+1)\big[2 - (5p+1)\big] \;=\;0[/tex]

. . . . . . [tex]\tfrac{1}{2}e^{-5p}(5p+1)(1-5p) \;=\;0[/tex]

Hence: .[tex]p = \tfrac{1}{5},\; \color{red}{\rlap{//////}}p = \text{-}\tfrac{1}{5}[/tex]Substitute into [1]: .[tex]A \;=\;\tfrac{1}{10}e^{-5(\frac{1}{5})}\left(5[\tfrac{1}{5}] + 1\right)^2 \;=\; \tfrac{1}{10}e^{-1}(4)[/tex]

Therefore: .[tex]\text{max }A \:=\:\frac{2}{5e}[/tex]

Beautifully latexed.

How can you be sure this is a maximum?

 

[Petrus]

Hello,
soroban your answer is correct. Honestly I could not find in My book about this(I Dont use same calculus book). I would like to read about it but I can't find in Google. Can someone tell me/Link me about this? One more question, can I Also solve this with anti derivate?

 

[MarkFL]

I would think your textbook should have a chapter about the applications of the derivative, and a section on extrema of functions.

Optimization is an application of differentiation, as this allows you to find where the function has zero slope, and relative extrema occur at such critical points.

 

[MarkFL]

Well I could make a formula \(\displaystyle y=-5e^{-5x}(x_1-x)-e^{-5x}\) is this correct? Where x is our point

That is close, however the point-slope formula is:

\(\displaystyle y-y_1=m(x-x_1)\)

where in our case, we have:

\(\displaystyle y_1=f(x_1)=e^{-5x_1}\)

\(\displaystyle m=f'(x_1)=-5e^{-5x_1}\)

Now, once you write the correct tangent line, I would suggest expressing it in the two intercept form \(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

where then the $x$-intercept is $(a,0)$ and the $y$-intercept is $(0,b)$.

Can you explain whether we need to be concerned about the signs of the intercepts, and why?

 

[Petrus]

That is close, however the point-slope formula is:

\(\displaystyle y-y_1=m(x-x_1)\)

where in our case, we have:

\(\displaystyle y_1=f(x_1)=e^{-5x_1}\)

\(\displaystyle m=f'(x_1)=-5e^{-5x_1}\)

Now, once you write the correct tangent line, I would suggest expressing it in the two intercept form \(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

where then the $x$-intercept is $(a,0)$ and the $y$-intercept is $(0,b)$.

Can you explain whether we need to be concerned about the signs of the intercepts, and why?

Hello Mark,
I would start replace that \(\displaystyle x_1\) with c because I always confused myself on paper with them...:/ so we got \(\displaystyle (c,e^{-5c})\)
So our tangent function become \(\displaystyle y=-5e^{-5c}(x-c)+e^{-5c}\)
If we calculate y intercept \(\displaystyle (0,b)\)
I get \(\displaystyle b=-5e^{-5c}(0-c)+e^{-5c}\)
If we calculate x intercept we get
\(\displaystyle e^{-5c}(-5(x-c)+1)\)
is this correct? I did not understand that 'two intercept form' and that question (I would like to if you could explain your question another way cause I did not understand it).

 

[MarkFL]

Using your method works just as well...I will explain both. First your method

To find the $x$-intercept, we let $y=0$ and solve for $x$:

\(\displaystyle 0=-5e^{-5c}(x-c)+e^{-5c}\)

Since \(\displaystyle e^{-5c}\ne0\) we may divide through by this factor to obtain:

\(\displaystyle 0=-5(x-c)+1\)

\(\displaystyle 5x-5c=1\)

\(\displaystyle x=\frac{5c+1}{5}\)

So the $x$-intercept is \(\displaystyle \left(\frac{5c+1}{5},0 \right)\)

Now, to find the $y$-intercept, we let $x=0$ and solve for $y$:

\(\displaystyle y=-5e^{-5c}(0-c)+e^{-5c}\)

\(\displaystyle y=5ce^{-5c}+e^{-5c}\)

\(\displaystyle y=e^{-5c}(5c+1)\)

and so the $y$-intercept is \(\displaystyle \left(0,e^{-5c}(5c+1) \right)\)

To use the two-intercept formula, we begin with the tangent line:

\(\displaystyle y=-5e^{-5c}(x-c)+e^{-5c}\)

and arrange in the form \(\displaystyle \frac{x}{a}+\frac{y}{b}=1\). See if you can derive this formula, the two-intercept formula, using the fact that you want to find the line passing through $(0,b)$ and $(a,0)$.

\(\displaystyle y=-5e^{-5c}x+5ce^{-5c}+e^{-5c}\)

\(\displaystyle 5e^{-5c}x+y=e^{-5c}(5c+1)\)

\(\displaystyle \frac{x}{\frac{5c+1}{5}}+\frac{y}{e^{-5c}(5c+1)}=1\)

So we know the intercepts are the same as we found with the first method. Now, the question remains, do we need to worry about the signs of the intercepts?

 

[Petrus]

Using your method works just as well...I will explain both. First your method

To find the $x$-intercept, we let $y=0$ and solve for $x$:

\(\displaystyle 0=-5e^{-5c}(x-c)+e^{-5c}\)

Since \(\displaystyle e^{-5c}\ne0\) we may divide through by this factor to obtain:

\(\displaystyle 0=-5(x-c)+1\)

\(\displaystyle 5x-5c=1\)

\(\displaystyle x=\frac{5c+1}{5}\)

So the $x$-intercept is \(\displaystyle \left(\frac{5c+1}{5},0 \right)\)

Now, to find the $y$-intercept, we let $x=0$ and solve for $y$:

\(\displaystyle y=-5e^{-5c}(0-c)+e^{-5c}\)

\(\displaystyle y=5ce^{-5c}+e^{-5c}\)

\(\displaystyle y=e^{-5c}(5c+1)\)

and so the $y$-intercept is \(\displaystyle \left(0,e^{-5c}(5c+1) \right)\)

To use the two-intercept formula, we begin with the tangent line:

\(\displaystyle y=-5e^{-5c}(x-c)+e^{-5c}\)

and arrange in the form \(\displaystyle \frac{x}{a}+\frac{y}{b}=1\). See if you can derive this formula, the two-intercept formula, using the fact that you want to find the line passing through $(0,b)$ and $(a,0)$.

\(\displaystyle y=-5e^{-5c}x+5ce^{-5c}+e^{-5c}\)

\(\displaystyle 5e^{-5c}x+y=e^{-5c}(5c+1)\)

\(\displaystyle \frac{x}{\frac{5c+1}{5}}+\frac{y}{e^{-5c}(5c+1)}=1\)

So we know the intercepts are the same as we found with the first method. Now, the question remains, do we need to worry about the signs of the intercepts?

Do you mean if i should worry if it is positive or negative, It should be positive cause area can't be negative:)

 

[MarkFL]

Right, area cannot be negative, so is there any way for product of the intercepts to be negative? That is, is there any way that the two intercepts can have opposite signs?

 

[Petrus]

Right, area cannot be negative, so is there any way for product of the intercepts to be negative? That is, is there any way that the two intercepts can have opposite signs?

If you got exempel power of 2. Like \(\displaystyle x^2\) what I mean is if we got like Two x exempel \(\displaystyle x•x\)

 

[MarkFL]

Write out the product of the two intercepts and see if you can explain why it is always non-negative. While this may seem trivial, I think it is good to look at such things. :D

 

[Petrus]

Write out the product of the two intercepts and see if you can explain why it is always non-negative. While this may seem trivial, I think it is good to look at such things. :D

What you mean? If we take \(\displaystyle -1*-1=1\) and \(\displaystyle 1+1=1\) two negative in multiplying will cancel each?

 

[MarkFL]

The product of the two intercepts is:

\(\displaystyle \frac{5c+1}{5}\cdot e^{-5c}(5c+1)=\frac{(5c+1)^2e^{-5c}}{5}\)

Since \(\displaystyle 0<e^{-5c}\) and \(\displaystyle 0\le(5c+1)^2\), this product is always non-negative. This way we are sure we need not worry about possibly getting a negative area. :D

 

[Petrus]

The product of the two intercepts is:

\(\displaystyle \frac{5c+1}{5}\cdot e^{-5c}(5c+1)=\frac{(5c+1)^2e^{-5c}}{5}\)

Since \(\displaystyle 0<e^{-5c}\) and \(\displaystyle 0\le(5c+1)^2\), this product is always non-negative. This way we are sure we need not worry about possibly getting a negative area. :D

Hello Mark,
I understand the first point because \(\displaystyle e^{-5c}=\frac{1}{e^{5c}}>0\) and the last part is \(\displaystyle 0\le(5c+1)^2\) because in the problem it says \(\displaystyle x>0\) but we did subsitute that x is c so \(\displaystyle c>0\) that means \(\displaystyle 0\le(5c+1)^2\) ,c ∈ ℕ Is this correct explain?

 

[MarkFL]

The square of a real value can never be negative. It's as simple as that. :D

 

[Petrus]

The square of a real value can never be negative. It's as simple as that. :D

ohh I wanted to explain really good ohh well :P Well now we got out intercept then I should rewrite the area?

 

[MarkFL]

Yes, now write the function representing the area of the described triangle, and use differentiation to find the maximum, and be able to demonstrate that it is a maximum.

 

[Petrus]

Hello Mark,
We put those x and y in area and derivate it and we get
\(\displaystyle (-5^{-5c}(5c+1))(-2+5c+1)=0\) so we get \(\displaystyle c=\frac{1}{5}\) and \(\displaystyle c_1=-\frac{1}{5}\) and we put it in on our Area function
\(\displaystyle A=\frac{e^{-5c}(5c+1)^2}{10}\) and if we put \(\displaystyle c=-\frac{1}{5}\) area will be 0. and if we put \(\displaystyle c=\frac{1}{5}\) we get the answer \(\displaystyle \frac{2}{5e}\) that means our max area is \(\displaystyle \frac{2}{5e}\)

 

[MarkFL]

You should actually get:

\(\displaystyle A'(c)=\frac{1}{2}e^{-5c}(5c+1)(1-5c)=0\)

However, you have gotten the correct critical values.

Additionally, I would suggest using one of the extrema tests (first or second derivative tests) to ensure you have indeed found the maximum. We can be fooled by the fact that the derivative is zero at a point into thinking it must be an extremum. We should verify it by one the the tests. (Wink)

 

[Petrus]

You should actually get:

\(\displaystyle A'(c)=\frac{1}{2}e^{-5c}(5c+1)(1-5c)=0\)

However, you have gotten the correct critical values.

Additionally, I would suggest using one of the extrema tests (first or second derivative tests) to ensure you have indeed found the maximum. We can be fooled by the fact that the derivative is zero at a point into thinking it must be an extremum. We should verify it by one the the tests. (Wink)

Well I did ignore that 1/2 huh I should not forget about it... I will do it later:)