Let
x------> the length side of the square base of the box
y-------> the height of the box
we know that
volume of the box=b²*h
b=x
h=y
volume=256 cm³
so
256=x²*y------>y=256/x²--------> equation 1
<span>The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.</span>
surface area of the cardboard=area of the base+perimeter of base*height
area of the base=x²
perimeter of the base=4*x
height=y
surface area=x²+4x*y-----> equation 2
substitute equation 1 in equation 2
SA=x²+4x*[256/x²]-----> SA=x²+1024/x
step 1
find the first derivative of SA and equate to zero
2x+1024*(-1)/x²=0------> 2x=1024/x²----> x³=512--------> x=8 cm
y=256/x²------> y=256/8²-----> y=4 cm
the answer is
the length side of the square base of the box is 8 cm
the height of the box is 4 cm
Answer: 5
15/3 = 5
There are 5 doughnuts left.
General Idea:
Domain of a function means the values of x which will give a DEFINED output for the function.
Applying the concept:
Given that the x represent the time in seconds, f(x) represent the height of food packet.
Time cannot be a negative value, so
The height of the food packet cannot be a negative value, so
We need to replace 
for f(x) in the above inequality to find the domain.
![-15x^2+6000\geq 0 \; \; [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0](https://tex.z-dn.net/?f=%20-15x%5E2%2B6000%5Cgeq%200%20%5C%3B%20%5C%3B%20%20%5BDivide%20%5C%3B%20by%5C%3B%20-15%5C%3B%20on%5C%3B%20both%5C%3B%20sides%5D%5C%5C%20%5C%5C%20%5Cfrac%7B-15x%5E2%7D%7B-15%7D%20%2B%5Cfrac%7B6000%7D%7B-15%7D%20%5Cleq%20%5Cfrac%7B0%7D%7B-15%7D%20%5C%5C%20%5C%5C%20x%5E2-400%5Cleq%200%5C%3B%5BFactoring%5C%3Bon%5C%3Bleft%5C%3Bside%5D%5C%5C%20%5C%5C%20%28x%2B200%29%28x-200%29%5Cleq%200%20)
The possible solutions of the above inequality are given by the intervals ![(-\infty , -2], [-2,2], [2,\infty )](https://tex.z-dn.net/?f=%20%28-%5Cinfty%20%2C%20-2%5D%2C%20%5B-2%2C2%5D%2C%20%5B2%2C%5Cinfty%20%29%20)
. We need to pick test point from each possible solution interval and check whether that test point make the inequality 
true. Only the test point from the solution interval [-200, 200] make the inequality true.
The values of x which will make the above inequality TRUE is 
But we already know x should be positive, because time cannot be negative.
Conclusion:
Domain of the given function is 
X = -6y - 12
4x + 5y =-39
4x + 5y = -39
-4(-6y - 12) + 5y = -39
-4(-6y) + 4(12) + 5y = -39
24y + 48 + 5y = -39
29y + 48 = -39
29y = -87
y = -3
x = -6y - 12
x = -6(-3) - 12
x = 28 - 12
x = 6
(x, y) = (6, -3)
It is the last equation. You square both sides.
