Derivative applications

Question

I'm a little bit unsure of how to do these and I'm hoping someone can help me out.

1. The dimensions of a rectangle are changing in such a way that the perimeter remains 24 inches. Show that when the area is 32 square inches, the area is either increasing or decreasing 4 times as fast as the length is increasing.

24=2x+2y, x+y=12... not sure where to go from here.

2. An object is projected vertically upward from the ground. Show that it takes the object the same amount of time to reach its maximum height as it takes for it to drop from that height back to the ground. Measure the height in meters.

Thanks!

Answer 1

Assuming $x$ denotes the length, we know that $x+y=12$. Now using this, write the area of the rectangle as a function of only $x$. I am sure you can do the rest from there.

Answer 2

For 1, we let the dimensions of the rectangle be $l$ and $w$. Then the area is $A=lw$. Given that the perimeter of the rectangle is 24, write $24=P=2l+2w$, or $l+w=12$. Rearrange this to obtain $w=12-l$, and substitute this into the area formula to get $A=l(12-l)=12l-l^2$. Now we can find the length of the rectangle when the area is 32 by solving $32=12l-l^2$. You should get 4 or 8. Choose either, as the other will be the width. Then we calculate derivatives: $$\frac{dA}{dt}=12\frac{dl}{dt}-2l\frac{dl}{dt}$$ Substituting in $l=4$, we obtain the desired result: $$\frac{dA}{dt}=4\frac{dl}{dt}$$ Note that if we chose 8 for the length, the width would be 4, and we would obtain: $$\frac{dA}{dt}=-4\frac{dl}{dt}$$