Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
Answer:
Step-by-step explanation:
This is the associative property for addition. In informal language, it says under addition, the order of the addition is not critical.
Answer:
C. x = 3/13
Step-by-step explanation:
Answer:
y=-2/5x-3
Step-by-step explanation:
the slope is m and the y intercept is b
To get the extrema, derive the function.
You get y' = 2x^-1/3 - 2.
Set this equal to zero, and you get x=0 as the location of a critical point.
Since you are on a closed interval [-1, 1], those points can also have an extrema.
Your min is right, but the max isn't at (1,1). At x=-1, you get y=5 (y = 3(-1)^2/3 -2(-1); (-1)^2/3 = 1, not -1).
Thus, the maximum is at (-1, 5).
