Question

Calculus: A window has the shape of an equilateral triangle of side 12ft. If a window blind is being lowered over the window at a rate of 0.5 feet per minute, how fast is the covered area of the window (shaded) increasing when the blind has been lowered 3 foot below the top of the window?

Answer 1

Welcome to Open Study! :)

Answer 2

It's increasing at a steady rate of 0.5 feet a minute as it states at the beginning of the question @nonards

Answer 3

the window blind is going down at that rate, but not the area that the triangle is trying to cover, which the question is asking for, but thank you and thank you for trying :)

Answer 4

oh ok I see sorry bout that if I'm correct you would have 9 ft left to cover up

Answer 5

Satellite, that is a pretty accurate drawing of the situation, shelby the sides are 12ft not the height

Answer 6

we need an expression for the area in terms of \(x\) i think, since what you know is \(x'=.5\)

Answer 7

if the height of the triangle i wrote above is \(x\) then the base is \(\frac{x}{\sqrt3}\) by trig or pythagaras or whatever
then the area is \(\frac{1}{2}x\times \frac{x}{\sqrt3}\) but since you have two of them those triangles the total area is \(\frac{x^2}{\sqrt3}\)

Answer 8

take the derivative wrt time and get \[A=\frac{x^2}{\sqrt3}\]
\[A'=\frac{2}{\sqrt3}xx'\]

Answer 9

you know \(x'=.5\) and you are told \(x=3\) so you can find \(A'\)
notice we did not use the \(12\) in this computation, so maybe the picture is wrong and it should look like this