A piece of cardboard measures 10 x 15 inches.

Question

A piece of cardboard measures 10 x 15 inches. Four equal squares are removed from corners of all sides. Find the maximum volume.

Answer 1

V = (10 - 2x)(15 - 2x)x = 4x³- 50x² + 150x; x > 0

V' = 12x² - 100x + 150 = 0, 6x² - 50x + 75 = 0; x = (25±5√7)/6

x = (25 - 5√7)/6 = 1.962 inch.

V ''(x) = 24x - 100; V''((25 - 5√7)/6) = 100 - 20√7 - 100 = - 20√7 < 0. So, we have maximum if x = (25 - 5√7)/6 =

1.962 inch

Answer 2

V = x(10-2x)(15-2x) = 4x^3 -50x^2 +150x

take the derivative and set equal to zero, then solve for x

V' = 12x^2 -100x + 150 = 0

divide by 4

3x^2 -25x + 37.5 = 0

x =25/6 + (1/6)sqr(625 - 12(37.5))

x= 4 1/6 + (1/6)sqr(175) = 4.67 - 3.54 = 1.13 inches

max V = 1.13(10-2.26)(15-2.26) = 1.13(7.74)(12.74) = 111.4 cubic inches

Answer 3

Well suppose the side length of the squares removed is s

Then if you fold up the new cardboard into a box, the height will be s, and the base will be (10-2s) x (15-2s)

So volume = s (10-2s) (15-2s)

You need to find s, to make this as large as possible

Plot the function on Desmos or a TI-84 and choose s to make the function as large as possible

(obviously 0 < s < 5).

Then when you have found the s, you can calculate the maximum volume