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Triangle max area

Petrus

Well-known member
Feb 21, 2013
739
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

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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

Well-known member
Feb 21, 2013
739
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. lets 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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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

Well-known member
Feb 2, 2012
409
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
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

Well-known member
Feb 21, 2013
739
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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 cant be negative:)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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 Is this correct explain?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The square of a real value can never be negative. It's as simple as that. :D
 

Petrus

Well-known member
Feb 21, 2013
739
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Feb 21, 2013
739
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 \(\displaystyle \frac{2}{5e}\)