A recatangle has constant area of 200 square meters and its lenght L is increasing at the rate of 4 meters per second

a) find the width W at the instant that the width is decreasing at the rate of .5 meters per second
b) at what rate is the diagonal d of the rectangle changing at the instant when the width is 10 meters?

A = 200 = W * L

dL/dt = 4

W = 200/L
dW/dL = -200/L^2
dW/dt = -(200/L^2)dL/dt

if dW/dt = -.5
then
-.5 = -(200/L^2) * 4
L^2 = 1600
L = 40
W = 200/40 = 5

go through that again for W = 10 to get L, dW/dt
then
D^2 = L^2 + W^2
2 D dD/dt = 2 L dL/dt + 2 W dW/dt

To solve these problems, we can use the formulas for the area of a rectangle, the Pythagorean theorem, and differentiation.

a) To find the width W at the instant that the width is decreasing at the rate of 0.5 meters per second, we need to relate the width and length of the rectangle. The area of a rectangle is given by the formula A = L * W, where A is the area, L is the length, and W is the width.

Given that the area is constant at 200 square meters, we have A = 200. We also know that dA/dt = 0 because the area is constant.

Taking the derivative of the area formula with respect to time, we get dA/dt = d(L * W)/dt. Applying the product rule, we have dA/dt = L * dW/dt + W * dL/dt.

Since dA/dt = 0, we can rewrite the equation as 0 = L * dW/dt + W * dL/dt.

Substituting the given values dL/dt = 4 m/s and dW/dt = -0.5 m/s (negative sign because the width is decreasing), we have 0 = L * (-0.5) + W * 4.

Since we are interested in finding the width W when the width is decreasing at 0.5 m/s, we can substitute dW/dt = -0.5 m/s and solve for W.

0 = L * (-0.5) + W * 4
0 = -0.5L + 4W
0.5L = 4W
W = 0.5L/4
W = 0.125L

Substituting the known value L = 10 (since it's increasing at 4 m/s), we can find W.

W = 0.125 * 10
W = 1.25 m

Therefore, when the width is decreasing at 0.5 m/s, the width of the rectangle is 1.25 meters.

b) To find the rate at which the diagonal d of the rectangle is changing when the width is 10 meters, we need to use the Pythagorean theorem.

The Pythagorean theorem states that the square of the diagonal is equal to the sum of the squares of the width and length. Mathematically, this can be written as d^2 = W^2 + L^2.

Differentiating both sides of the equation with respect to time, we get 2d * dd/dt = 2W * dW/dt + 2L * dL/dt.

Since we are interested in finding the rate of change of the diagonal when the width is 10 meters, we can substitute W = 10 into the equation.

2d * dd/dt = 2(10) * dW/dt + 2L * dL/dt

Given that dW/dt = 0 (since the width is constant) and dL/dt = 4 m/s (since the length is increasing at 4 m/s), we have:

2d * dd/dt = 2(10) * 0 + 2(10) * 4
2d * dd/dt = 80

Dividing both sides by 2d, we get:

dd/dt = 80 / (2d)
dd/dt = 40 / d

To find the rate at which the diagonal is changing when the width is 10 meters, we need to substitute d = sqrt(W^2 + L^2) = sqrt(10^2 + 10^2) = sqrt(200) = 10√2 into the equation.

dd/dt = 40 / (10√2)
dd/dt = 4 / √2
dd/dt = 4√2 / 2
dd/dt = 2√2 m/s

Therefore, at the instant when the width is 10 meters, the diagonal of the rectangle is changing at a rate of 2√2 meters per second.

To solve these problems, we will use the formulas for the area, width, and diagonal of a rectangle.

a) Finding the width W when it is decreasing at a certain rate:
The formula for the area of a rectangle is given by A = L * W, where A is the area, L is the length, and W is the width.

Since the area is constant at 200 square meters, we can rewrite the formula as 200 = L * W.

To find the rate at which the width is changing, we need to take the derivative of both sides of the equation with respect to time t.

d/dt (200) = d/dt (L * W)

0 = dL/dt * W + L * dW/dt

Since the length L is increasing at a rate of 4 meters per second (dL/dt = 4), and the width is decreasing at a rate of -0.5 meters per second (dW/dt = -0.5), plugging these values into the equation gives us:

0 = 4 * W + L * (-0.5)

0 = 4W - 0.5L

Now we substitute the value of L from the original equation (200 = L * W) into the equation above:

0 = 4W - 0.5 * (200 / W)

Simplifying, we get:

0 = 4W - 100 / W

Multiplying everything by W to eliminate the denominator, we have:

0 = 4W^2 - 100

Rearranging the equation, we get a quadratic equation:

4W^2 - 100 = 0

Solving this quadratic equation will give us the values of W. In this case, we are looking for the width W when it is decreasing, so we'll select the negative value.

b) Finding the rate of change of the diagonal d when the width is 10 meters:

The formula for the diagonal of a rectangle is given by the Pythagorean theorem: d^2 = L^2 + W^2, where d is the diagonal, L is the length, and W is the width.

Taking the derivative of both sides of the equation with respect to time t, we get:

2d * (dd/dt) = 2L * (dL/dt) + 2W * (dW/dt)

Since we are interested in finding the rate of change of the diagonal when the width is 10 meters, we can substitute the values into the equation. Let's assume the length L remains constant.

2d * (dd/dt) = 2L * 0 + 2(10) * (dW/dt)

Simplifying, we get:

2d * (dd/dt) = 20 * (dW/dt)

Now we can solve for (dd/dt) by dividing both sides of the equation by 2d:

(dd/dt) = 10 * (dW/dt) / d

Substitute the given value of the width (W = 10) into the equation to find the rate of change of the diagonal at that instant.