Let x represent the side length of the square end, and let d represent the dimension that is the sum of length and girth. Then the volume V is given by
V = x²(d -4x)
Volume will be maximized when the derivative of V is zero.
dV/dx = 0 = -12x² +2dx
0 = -2x(6x -d)
This has solutions
x = 0, x = d/6
a) The largest possible volume is
(d/6)²(d -4d/6) = 2(d/6)³
= 2(108 in/6)³ = 11,664 in³
b) The dimensions of the package with largest volume are
d/6 = 18 inches square by
d -4d/6 = d/3 = 36 inches long
Answer:
2iiej
Step-by-step explanation:
1kejes191828wusnalao
Answer:
11.33 * 16.33 meters to the nearest hundredth.
Step-by-step explanation:
Let the width of the pool be x meters, then the length is x+5 meters.
The length of whole area = x + 5 + 2(3) = x + 11 meters and the width is
x + 2(3) = x + 6 meters.
So we have the equation (x + 6)(x + 11) = 387.
x^2 + 17x + 66 = 387
x^2 + 17x - 321 = 0
x = [-17 +/- sqrt (17^2 - 4*1*-321)] / 2
x = 11.33 meters
So the width is 11.3 m and the length is 16.33 meters.