This question was given to Mrs. RickJay’s math course; these students were just taught the concept of derivatives four weeks ago.
“A square of side X is cut from each corner of a square sheet of metal 16 cm on a side, and the edges are turned up to form a tray. Write an equation for the volume of the tray and determine the value of X which will yield the maximum volume.”
Man, this is INTRODUCTORY math?
Anyway, I am pretty sure the equation for volume is V=(16-x)(16-x)(x), right? In that case, how would I find the max value?
That’s easy. First, there’s a turnup on each side, so the sides are (16-2x). Other than that, your equation is fine, but you need to muliply it out.
If she learned about derivatives a month ago, she should know what happens to a function’s derivative at the function’s maximum. That’s all the hints I’m gonna give.
No, the formula for the volume would be (16-2x)(16-2x)(x). Remember each side has 2 cuts of length x on it. I’m not sure of the maximum volume condition though.
I believe the volume is (16-2x)(16-2x)(x). You are cutting a square of side x from each corner, so x is taken away twice from each side. Pretend x = 2. By cutting the upper right hand corner, the top length is now 14. Cutting the upper left corner makes it 12.
Maximize the volume by multiplying out the equation and taking the derivative. Set it equal to 0 and solve for x. If I have done this right, x is between 1 and 2, but further deponent sayeth not. Anyone wanna check me out on this?
The volume would be given by:
V = x * (16 - 2x)^2 (where 0 < x < 8)
Solving the problem by finding the place where the derivative of this function is zero.
This is what’s known as an optimization problem, and it’s one of the only things that Calculus is good for. So you’ll have to expect to see it a lot in a course on differential Calculus. You’re right that you need to identify V(x) and find the maximum value of V(x) for x = 0 to 8 (the possible range). To maximize a function, you set the derivative equal to 0, solve for x, and plug it back in. Here’s an example:
What’s the maximum value of f(x) = -x[sup]2[/sup] + 6x?
First, set the derivative equal to 0 and solve for x:
df/dx = 0
-2x + 6 = 0
x = 3
Now, plug it back in to f(x): The maximum value is f(3) = 9.
(I left something out of here; you don’t actually know if that’s a maximum or a minimum. There are a number of good ways to tell, but I don’t expect that the class has gotten there yet.)
You’re not far off, but that’s not exactly why I did a different equation. I genuinely think that providing another example is more instructive than doing the problem as posed. I guess I should have made it clear, though, that I was solving a different equation.
