A particular piece of cardboard has a length of 20cm and a width of 16cm. The side lnegth of a corner square is x.
a.) Derive a polynomial function that represents the volume of the box?
b.) What is an appropriate domain for the volume function?
d.) What should be the length of a corner square if the volume of the box is maximized?
e.) For what values of x is the volume of the box greater than 200cm^3?
- Wayne DeguManLv 72 months agoFavorite Answer
Removing squares of side x cm will leave a tray with dimensions 20 - 2x by 16 - 2x by x
so, volume => x(20 - 2x)(16 - 2x)
i.e. V = 4x(10 - x)(8 - x)
Domain --> 0 < x < 8
Now, V = 4x³ - 72x² + 320x
so, dV/dx = 12x² - 144x + 320 = 0....at maximum
Using the quadratic formula we get:
x = 2.94 or 9.06
so, fitting with the domain, x = 2.94
Hence, Vmax = 4(2.94)³ - 72(2.94)² + 320(2.94)
i.e. 420.1 cm³
Now, d²V/dx² = 24x - 144
With x = 2.94, d²V/dx² < 0....i.e. maximum
For the volume to be greater than 200 cm³ we require,
4x³ - 72x² + 320x > 200
i.e. 4x³ - 72x² + 320x - 200 > 0
Using numerical techniques we can say:
0.75 < x < 5.93
A sketch is below.
- AmyLv 72 months ago
I assume this is the "cut out the corners and fold the sides up" problem. You can google it to find many, many worked examples already out there.
(a) volume = length * width * height. Draw a picture of the cutout cardboard if you're having trouble visualizing how x affects the length, width, and height.
(b) since you're dealing with a physical box, x cannot be negative and two corners cannot exceed the width of the box.
(d) Take your function of x from part (a). Set the derivative equal to zero. Solve. Identify whether each solution is a minimum or maximum or neither.
(e) Take your function of x from part (a). Set it equal to 200cm^3. That gives you the endpoints for intervals in which the volume is either always greater than 200 or always less than 200. Check any number from inside each interval to see which.