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
(x + 3)^2 + (x + 4)^2
= x^2 + 6x + 9 + x^2 + 8x + 16
= 2x^2 + 14x + 25
= 2(x^2 + 7x) + 25
= 2[(x + 7/2)^2 - 49/4] + 25
= 2(x + 7/2)^2 - 98/4 + 25
= 2(x + 7/2)^2 + 1/2
Its B
Answer:
C) (x-3)(x+3)
Note:
Just a tip there is an app called Desmos where you can put in equations like this and it will graph them for you.
Answer:
The answer is three please give brainliest:D
Step-by-step explanation:
