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:
5x - 2y - 23 = 0
Step-by-step explanation:
Line is passing through the points 
Equation of line in two point form is given as:
The answer would be: D-100. The reasoning of this is because if you take 3 X 10 to the third power and did the exponent (10 to the third power in this case) and multiplied it you would obviously get 1,000 for an answer. Then would would do 1,000 X 3 to get 3,000.
5X10=50 sq ft area
Around the 3 sides= 5+10+5=20
