Area of a triangle under a curve

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SUMMARY

The discussion focuses on maximizing the area of a triangle formed by the tangent line of the curve y=e^(-x) and the axes, with the origin O as one vertex. The area formula used is A = (1/2) * width * height, where width and height are expressed in terms of x_0. The derivative of the area function is calculated as d/dx_0((e^(-x_0))*(x_0 + 1)^2) to find the maximum area. The solution confirms the correct approach to derive the maximum area of the triangle.

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  • Understanding of calculus, specifically derivatives
  • Familiarity with the exponential function y=e^(-x)
  • Knowledge of tangent lines and their equations
  • Basic geometry concepts related to triangles and area calculation
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  • Explore the application of the tangent line in optimization problems
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diredragon
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Homework Statement


The tangent line of a curve y=e^(-x) intercepts the axises at points A and B. What is the maximum area of a triangle AOB considering O as the origin.

Homework Equations


Ar= xy/2

The Attempt at a Solution


Derivative of this function is y'=-e^(-x)
I took the formula of the tangent line
y - yo = -e^(-x)(x-xo) and solved for x=0 and y=0 getting two equations
y = (xo + 1)yo and x = 1+ xo yet i don't know where to follow from this
 
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You have the width and height of the triangle in terms of x_0. That then gives you the area in terms of x_0, which you can maximize.
 
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diredragon said:
The tangent line of a curve y=e^(-x) intercepts the axises at points A and B.
Side note -- "axises" is not a word in English. The plural of "axis" is "axes".
One axis, two axes.
 
pasmith said:
You have the width and height of the triangle in terms of x_0. That then gives you the area in terms of x_0, which you can maximize.
So i take (d/dxo)((e^(-xo))*(xo + 1)^2) and whatever i get is the value of the maximum area right?
 
Solved it. Thanks!
 

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