A 13-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away at a rate of 2 feet per second, how fast is the area of the triangle formed by the wall, the ground, and the ladder changing when the bottom is 12 feet away from the wall?
Rice production requires both labor and capital investments in equipment and land. Suppose that if x dollars per acre aer invested in labor and y dollars per acre are investd in equpiment and land, then the yield P of rice per acre is given by the formula P = 100 times the square root of x + a50 times the square root of y. If a farmer invests $40/acre, how hould he divide the $40 between labor and capital investment in order to maximize the amount produced?
To solve this problem, we can use the concept of related rates. We need to find the rate of change of the area of the triangle with respect to time.
Let's break this problem down step by step:
1. First, let's label the variables in the problem:
- The height of the ladder is given as 13 feet.
- The distance between the bottom of the ladder and the wall is changing, and it's given as 12 feet.
2. We need to find the rate of change of the area of the triangle, which we'll call "A," with respect to time, which we'll call "t." Therefore, we want to find dA/dt.
3. The area of a triangle can be calculated using the formula A = (1/2) * base * height. In this case, the base of the triangle is the distance between the wall and the bottom of the ladder (which is changing), and the height is the length of the ladder (which is constant).
4. Now, let's determine the relationship between the changing quantities. We can use the Pythagorean theorem to relate the base and height of the triangle:
- The ladder, base, and height form a right triangle, so by the Pythagorean theorem, we have: base^2 + height^2 = ladder^2.
- Squaring both sides, we get: base^2 = ladder^2 - height^2.
5. Since the height is constant (13 feet), we need to express the base in terms of the changing quantity: the distance between the wall and the bottom of the ladder. Let's call this distance "x."
- From step 4, we have: base^2 = 13^2 - x^2.
- Taking the square root of both sides, we get: base = sqrt(13^2 - x^2).
6. Now, we can differentiate both sides of the equation with respect to time (t), remembering to use the chain rule:
- d(base)/dt = d(sqrt(13^2 - x^2))/dt.
7. Let's evaluate each term on the right side:
- The derivative of sqrt(13^2 - x^2) with respect to t is:
- (1/2)*(13^2 - x^2)^(-1/2)*(-2x)*dx/dt.
- As given in the problem, dx/dt = -2 ft/s (since the bottom of the ladder is being pulled away at a rate of 2 ft/s).
8. Substituting the known values, we have:
- d(base)/dt = (1/2)*(13^2 - x^2)^(-1/2)*(-2x)*(-2).
- Simplifying, we get d(base)/dt = x/sqrt(13^2 - x^2).
9. Now, we can substitute this value into the formula for the area of the triangle: A = (1/2) * base * height.
- A = (1/2) * (x/sqrt(13^2 - x^2)) * 13.
10. Finally, we can differentiate the equation for the area of the triangle with respect to time (t) to find dA/dt:
- dA/dt = (1/2) * (1/sqrt(13^2 - x^2)) * (dx/dt) * 13.
11. Substituting the known values, we have:
- dA/dt = (1/2) * (1/sqrt(13^2 - x^2)) * (-2) * 13.
12. Simplifying the equation, we get:
- dA/dt = -13(x/sqrt(13^2 - x^2)).
Now, we can calculate the value of dA/dt when x = 12 feet to find how fast the area of the triangle is changing when the bottom is 12 feet away from the wall.
Let h be the height of the ladder against the wall. Let x be the distance of the base of the ladder from the wall.
At the time specified, you have a good old 5-12-13 triangle.
x^2 + h^2 = 13^2
2x dx + 2h dh = 0
2(12)(2) + 2(5)dh = 0
dh/dt = -24/5 = -4.8
a = 1/2 xh
da = 1/2 (x dh + h dx)
da = 1/2 (-12*4.8 + 5*2) = -23.8 ft^2/sec