Abox (with no top) will be made by cutting squares of equal size out of the corners of a 2626 inch by 4343 inch rectangular piece of cardboard, then folding the side flaps up. find the maximum volume of such a box.

If x represents the side length of the square cut from each corner, then the dimensions of the base of the box are 26-2x by 43-2x. The volume of the box is

... V = x(26-2x)(43-2x)

where x is restricted to the domain 0 ≤ x ≤ 13.

Plotting this equation using a graphing calculator shows the maximum volume to be 2644.7 in³.

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The analytical solution can be found by setting the derivative of V to zero.

... V = 4x³ - 138x² +1118x

... dV/dx = 0 = 12x² -276x +1118

This has solutions given by the quadratic formula, where a=12, b=-276, c=1118:

... x = (-b±√(b²-4ac))/(2a)

The plus sign will give a solution that is not in the allowed domain of x, so the only viable solution is the one where the radical is subtracted.

... x = (276 - √(76176--53664))/24 = (23/2) - √(469/12) ≈ 5.2483 . . . inches

Then the corresponding maximum volume is found from the equation for V.

... V ≈ (5.2483)(26 - 2·5.2483)(43 - 2·5.2483) ≈ 2644.6877 in³ . . . . as above

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.