You are planning to close off the 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

From applications of derivatives...this is really tough!!

Write the equation that relates a to b:
a^2 + b^2 = 20^2 = 400
Now differentiate with respect to a, with b as a function of a.
2 a + 2 b db/da = 0
db/da = -(a/b)

Area = a b/2
Call the area A
dA/da = 0 at maximum
b + a db/da = 0
b + a (-a/b) = 0
a^2/b = ba^2 = b^2
Therefore a = b, since both are positive

To show that the area of the triangle enclosed by the segment is largest when a=b, we need to find the critical points of the area equation A = (1/2)ab and determine whether they correspond to maximum or minimum values.

First, we know that the equation relating a and b is a^2 + b^2 = 400 (from a^2 + b^2 = 20^2 = 400).

Differentiating this equation with respect to a, we get:
2a + 2b * (db/da) = 0.

Simplifying, we have:
db/da = -a/b.

Now, we can express the area A in terms of a:
A = (1/2)ab.

To find the critical points, we take the derivative of A with respect to a and set it equal to zero:
dA/da = (1/2) * [(da/da)b + a*(db/da)] = 0.

Substituting the value of db/da (-a/b) from above, we have:
(1/2) * (b - a^2/b) = 0.

Multiplying through by b gives:
(1/2)b^2 - (1/2)a^2 = 0.

Rearranging the equation, we have:
(1/2)b^2 = (1/2)a^2.

Since both a and b are positive, we can cancel the common factor of (1/2) from both sides to get:
b^2 = a^2.

Taking the square root of both sides, we find:
b = a.

Therefore, we have shown that the area of the triangle is largest when a=b.

To show that the area of the triangle enclosed by the line segment is largest when a = b, we can use the method of optimization by finding the critical points of the area equation.

First, let's write the equation that relates a and b using the Pythagorean theorem. The length of the line segment is 20 units, so we have:

a^2 + b^2 = 20^2 = 400

Now, let's differentiate both sides of the equation with respect to a, keeping in mind that b is a function of a:

2a + 2b * (db/da) = 0

We want to find the value of a where the rate of change of b with respect to a is 0, so we solve for db/da:

db/da = -(a/b)

Now, let's find the area of the triangle enclosed by the line segment. The formula for the area of a triangle is (1/2) * base * height, and in this case, the base is a and the height is b. So:

Area = (1/2) * a * b

Let's call the area A. To find the critical points of the area equation, we differentiate it with respect to a and set the result equal to 0:

dA/da = 0

Using the area formula, we have:

db/da + a * (db/da) = 0

Substituting -(a/b) for db/da, we get:

-(a/b) + a * (-(a/b)) = 0

Simplifying the equation, we have:

a^2/b = ba^2 = b^2

Since a and b are both positive, we can conclude that a = b.

Therefore, the area of the triangle enclosed by the line segment is largest when a = b.

a line through ( 0,13) is 5 units from the origin in quadrant 1. find the equation.

please help me thanks!