optimzation problem # 10

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August 15th 2009, 06:07 PM
skeske1234

optimzation problem # 10

a rectangle is inscribed under the curve y=e^(-x)^2, with its base along the x-axis. Find the rectangle of largest area, subject to the restriction that the base does not exceed 4 units.

Now I want to maximize the area of the rectangle, so should I do this?

A=xy
=[e^(-x)^2]x

Tested the domain

f(0)=1
f(4)=0.0000000112

now this doesn't make much sense because we dont have a rectangle if the largest area is 1..... x =0? not rectangle.

Please guide me to enlightenment! really appreciate it. thank you
August 15th 2009, 09:50 PM
earboth

Quote:

Originally Posted by skeske1234

a rectangle is inscribed under the curve y=e^(-x)^2, with its base along the x-axis. Find the rectangle of largest area, subject to the restriction that the base does not exceed 4 units.

Now I want to maximize the area of the rectangle, so should I do this?

A=xy <<<<<<<< according to your calculations this is only the half of the area of the rectangle
=[e^(-x)^2]x

...

$A(x)=2x \cdot e^{-x^2}$

Calculate the first derivation of A. Use product rule:

$A'(x)=e^{-x^2} \cdot 2 + 2x \cdot e^{-x^2} \cdot (-2x)$

$A'(x)=2e^{-\frac12} \cdot (1-2x^2)$

Solve $A'(x)=0$ . You should come out with $x = \dfrac12 \cdot \sqrt{2}$

Plug in this value into the equation of A to get $A_{max}=\sqrt{2} \cdot e^{-\frac12} \approx 0.85776...$
August 16th 2009, 06:22 AM
skeske1234

Quote:

Originally Posted by earboth

$A(x)=2x \cdot e^{-x^2}$

Calculate the first derivation of A. Use product rule:

$A'(x)=e^{-x^2} \cdot 2 + 2x \cdot e^{-x^2} \cdot (-2x)$

$A'(x)=2e^{-\frac12} \cdot (1-2x^2)$

Solve $A'(x)=0$ . You should come out with $x = \dfrac12 \cdot \sqrt{2}$

Plug in this value into the equation of A to get $A_{max}=\sqrt{2} \cdot e^{-\frac12} \approx 0.85776...$

why is the area of the rectangle

A=2xy??
How is A=xy only half of the rectangle?
August 16th 2009, 07:05 AM
songoku

Hi skeske1234

If you draw the graph of $y=e^{-x^2}$ , it will be on 1st and 2nd quadrant and is symmetry about y-axis. So, the rectangle will also lie on 1st and 2nd quadrant and is symmetry about y-axis.

A = xy is only the area on one quadrant. If you take positive value of x, then the rectangle is on 1st quadrant. If you take negative value of x, the rectangle is on 2nd quadrant.

Hence, to get the total area : A = 2xy

Maybe, it will be clearer if someone posts the graph :)
August 16th 2009, 08:28 AM
skeeter

always sketch a graph before doing a problem like this ...
August 16th 2009, 10:12 PM
earboth

Quote:

Originally Posted by skeske1234

why is the area of the rectangle

A=2xy??
How is A=xy only half of the rectangle?

Have a look here: http://www.mathhelpforum.com/math-he...oblem-2-a.html