Posted by **Maria** on Sunday, November 13, 2011 at 10:23pm.

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?

- Calculus--Please Help! -
**Navjot**, Sunday, November 13, 2011 at 11:44pm
6seconds

- Calculus--Please Help! -
**htoo**, Monday, November 14, 2011 at 1:09am
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?

- Calculus--Please Help! -
**Steve**, Monday, November 14, 2011 at 9:15pm
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

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