Answer:
3. SSS
4. True
5. False
Step-by-step explanation:
Answer:
a b c
plz give branliest
Step-by-step explanation:
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
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Answer:
Step-by-step explanation:
Base on my own understanding what should be done is that three stacks will not work, because it wont be a symmetric arrangement and also one will be left out. So, two stacks of five each would be better and easy to carry. But the stacks should be arranged in such a manner that the lengths will be parallel to each other and not in-line which would increase the length making it comparatively very long. its easier to hold a (2*8.5,11,2*5=17,11,17) compact box, because it will be easy to carry a long and not heavy.
The answer is 0.7 Have a nice day