Wednesday
May 22, 2013

# Homework Help: largest possible area

Posted by Jen on Saturday, November 18, 2006 at 8:56pm.

Again from applications of derivatives..

what is the largest possible area for a right triangle whose hypotneus is 5cm long, and what are its dimensions.
Thanks

a^2 + b^2 = c^2

3^2 + 4^2 = 5^2

Area of a triangle
1/2 base times altitude

hint:
base is 3
altitude is 4

The formula for the area is:

A = 1/2 a*b (1)

where a and b are the two sides that meet at the right angle. The hypotenuse is sqrt[a^2 + b^2].

So, you must maximize A while keeping a^2 + b^2 fixed:

a^2 + b^2 = 25cm^2 (2)

There are three ways to solve such problems. The first way is what you learn in school and is straightforward but sometimes tedious method of eliminating variables from the constraints and finding the maximum for the remaining free variables.

In this case you have two variables (a and b), and 1 constraint. You can use the constraint to eliminate, say, b:

b = sqrt[25 cm^2 - a^2]

You can plug this equation into the formula for A in Eq. (1)

A = 1/2 a*b =

1/2 a*sqrt[25 cm^2 - a^2]

The next step is then to differentiate this w.r.t. a and set the derivative equal to zero.

The second method is to find a clever parametrization that will automatically satisfy the constraint. In this case you can put:

a = h cos(theta)

b = h sin(theta)

where h is the length of the hypotenuse and is equal to 5 cm.

Then A becomes:

1/2 h^2 cos(theta)*sin(theta)=

1/4 h^2 sin(2 theta).

Finding the maximum of this function is much easier!

A third method is to use the method of Lagrange multipliers. This is convenient when you can't solve the constraint equations or find suitable parametrizations that satisfy the contraints. According to this metod you must equate the derivatives of the function you wish to extremize to a linear combination of the derivatives of the constraint functions.

In this case it works as follows. If you put

g(a,b) = a^2 + b^2, then

dA/da = lambda dg/da

and

dA/db = lambda dg/db

where the derivatives are so-called partial derivatives, you evaluate them by keeping the other variables constant and lambda is an undetermined constant that you can fix later using the constraint

So, you get the equations:

1/2 b = lambda*2a

1/2 a = lambda*2b

If you divide these two equations you find that

b/a = a/b --->

a = b

Thank you.

I followed the first method and set the derivative of A equal to 0. I got the following answers for a

a = +5,-5, +5/sqrt(2), -5/sqrt(2)

Am I right? What's next?

I don't think the values of +/- 5 are correct. The other values do look ok. You need to take the positive value so, it's a = 5/sqrt(2). You then plug that into the equation for A.

You could have saved some work by maximizing A^2 instead of A. Because if A is at it's maximum, so is A^2 and vice versa.

The function A^2 is proportional to

a^2[25 cm^2 - a^2]

You can then put a^2 = x and consider this as a function of x:

x[25 cm^2 -x]

and maximize this as a function of x.

You find x = 25 cm^2/2 and thus a = 5cm/sqrt(2)

Thank you . I got it.

• largest possible area - J, Tuesday, January 5, 2010 at 2:19pm

The largest possible area is therefore A(max)=(1/2)xy =(1/2)[[5/sqrt(2)][5/sqrt(2)]] = (1/2)(25/2)= 25/4 sq cm

• largest possible area - Anonymous, Monday, May 16, 2011 at 1:23am

A= 6

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