It’s the third one.. I think
Your final answer is going to be 4r + 3r ^2 +7
The maximum volume of the box is 40√(10/27) cu in.
Here we see that volume is to be maximized
The surface area of the box is 40 sq in
Since the top lid is open, the surface area will be
lb + 2lh + 2bh = 40
Now, the length is equal to the breadth.
Let them be x in
Hence,
x² + 2xh + 2xh = 40
or, 4xh = 40 - x²
or, h = 10/x - x/4
Let f(x) = volume of the box
= lbh
Hence,
f(x) = x²(10/x - x/4)
= 10x - x³/4
differentiating with respect to x and equating it to 0 gives us
f'(x) = 10 - 3x²/4 = 0
or, 3x²/4 = 10
or, x² = 40/3
Hence x will be equal to 2√(10/3)
Now to check whether this value of x will give us the max volume, we will find
f"(2√(10/3))
f"(x) = -3x/2
hence,
f"(2√(10/3)) = -3√(10/3)
Since the above value is negative, volume is maximum for x = 2√(10/3)
Hence volume
= 10 X 2√(10/3) - [2√(10/3)]³/4
= 2√(10/3) [10 - 10/3]
= 2√(10/3) X 20/3
= 40√(10/27) cu in
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Complete Question
(Image Attached)
Answer:
The width is 3 yds
Step-by-step explanation:
Area = length * width
18 = 6*w
Divide each side by 6
18/6 = 6w/6
3=w
The width is 3 yds
Which only lists multiples of 16? O1,2,4, 8, 16 O 16, 24, 32, 40 O16, 32, 48, 64 O 1,2, 4, 8, 12, 16
schepotkina [342]
Answer:
48 & 68
Step-by-step explanation:
if you multiply the numbers you will see that you get 48 & 68 multiple times