Optimization: small plane passing overhead; open-top box

[Seraphless]

I have two questions for all of you that I am struggling with:

#1. An observer is tracking a small plane flying at an altitude of 5000 ft. The plane flies directly over the observer on a horizontal path at the fixed rate of 1000 ft/min. Find the rate of change of the distance from the plane to the observer when the plane has flown 12,000 feet after passing directly over the observer.

On this first one I have:
the triangle rule, the 5-12-13 triangle?
I am clueless about how to pull equations out of this for differentiating...

#2. An -open- rectangular box with square base is to be made from 48 sq ft of material. What dimensions will result in a box with the largest possible volume?

On this second one I have:
v = yx[sup:gvz7qbfq]2[/sup:gvz7qbfq]
AreaofBase = x[sup:gvz7qbfq]2[/sup:gvz7qbfq]
AreaofSide = 4xy
4xy + x[sup:gvz7qbfq]2[/sup:gvz7qbfq] = 48

so substituting (x[sup:gvz7qbfq]2[/sup:gvz7qbfq])(48 - x[sup:gvz7qbfq]2[/sup:gvz7qbfq]/4x) into y in the volume equation would yield the volume answer which the derivitive of needs to be found. Do I need the quotient rule?

 

[Dr. Flim-Flam]

Re: Optimization

# 1 is a related rate problem. We are given the plane is 5,000 ft above the observer, dy/dt = 1,000 ft/min and at what rate is the plane leaving the observer when it has flown 12,000 ft by. Find dh/dt.

Ergo h^2 = 5000 + y^2, h(dh/dt) = y(dy/dt) = 12,000*1,000 = 12,000,000.

h = sqrt(5,000^2+12,000^2) = 13,000. 13,000(dh/dt) = 12,000,000. dh/dt = 12,000,000/13,000 = 923 ft/min.

 

[Dr. Flim-Flam]

Re: Optimization

# 2. You done most of it. V(x) =[x(48-x^2)]/4, V ' (x) = (1/4)(48-3x^2), setting slope equal to zero, we have x =4.

Substituting into y = (48-x^2)/4x = 2. Hence max volumn = V(4) = 32 cuft.