Volume
of a rectangular box = length x width x height<span>
From the problem statement,
length = 60 - 2x
width = 10 - 2x
height = x</span>
<span>
where x is the height of the box or the side of the equal squares from each
corner and turning up the sides
V = (60-2x) (10-2x) (x)
V = (60 - 2x) (10x - 2x^2)
V = 600x - 120x^2 -20x^2 + 4x^3
V = 4x^3 - 100x^2 + 600x
To maximize the volume, we differentiate the expression of the volume and
equate it to zero.
V = </span>4x^3 - 100x^2 + 600x<span>
dV/dx = 12x^2 - 200x + 600
12x^2 - 200x + 600 = 0</span>
<span>x^2 - 50/3x + 50 = 0
Solving for x,
x1 = 12.74 ; Volume = -315.56 (cannot be negative)
x2 = 3.92 ;
Volume = 1056.31So, the answer would be that the maximum volume would be 1056.31 cm^3.</span><span>
</span>
Step-by-step explanation:
3 and 4.
because the sum of the smaller two sides of a triangle have to be less than or equal to the 3rd larger side.
Another way to write this is:
x = 23y + 5
xy = 6732
Now plug in the first equation into the second:
(23y + 5)y = 6732
23y^2 + 5y - 6732 = 0
Either use quadratic equation or factor:
(y - 17)(23y + 396) = 0
y = 17 or -396/23
You know you can automatically eliminate the second y because it's negative and you need the two integers to multiply to a positive number (6732).
Plug y = 17 back into either equation (second might be easier):
17x = 6732
x = 396
18cm is the answer of this question
