Show that the area of the triangle enclosed by the segment is largest when a = b?

You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a,0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

The Area of the triangle is A = (1/2)ab

We know that a and b are related by a^2 + b^2 = 20^2
Use this to eliminate a from the area equation since
a^2 = 400-b^2 or a = sqrt(400-b^2)

So A = (1/2)* [sqrt(400-b^2)]*b
Now we want the Area A to be a max. So take the derivative dA/db and set equal to zero:
A ‘(b) = (1/2)*{ sqrt(400-b^2) + b*(1/2)(-2b)/sqrt(400-b^2) }
using the product and chain rule.
Set this = 0 and solve for b: you get
b = 20/sqrt(2)
then a = sqrt(400-b^2) = sqrt(400/2) = 20/sqrt(2)

Therefore, a = b for the maximum area.

(I’m assuming you’re in first year single variable calculus; in multivariable calculus, this can be done even quicker by other methods)

This entry was posted on Thursday, January 21st, 2010 at 11:31 pm and is filed under triangle area. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Show that the area of the triangle enclosed by the segment is largest when a = b?”

Duke on January 22nd, 2010 at 5:13 am

The Area of the triangle is A = (1/2)ab

We know that a and b are related by a^2 + b^2 = 20^2
Use this to eliminate a from the area equation since
a^2 = 400-b^2 or a = sqrt(400-b^2)

So A = (1/2)* [sqrt(400-b^2)]*b
Now we want the Area A to be a max. So take the derivative dA/db and set equal to zero:
A ‘(b) = (1/2)*{ sqrt(400-b^2) + b*(1/2)(-2b)/sqrt(400-b^2) }
using the product and chain rule.
Set this = 0 and solve for b: you get
b = 20/sqrt(2)
then a = sqrt(400-b^2) = sqrt(400/2) = 20/sqrt(2)

Therefore, a = b for the maximum area.

(I’m assuming you’re in first year single variable calculus; in multivariable calculus, this can be done even quicker by other methods)
References :

Area List