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:
SA = 967.6 cm²
Step-by-step explanation:
Surface Area of cylinder = 
<u><em>Where r = 7 cm, h = 15 cm</em></u>
=> SA = 2(3.14)(7)(15)+2(3.14)(7)²
=> SA = 659.7 + 2(3.14)(49)
=> SA = 659.7 + 307.9
=> SA = 967.6 cm²
Simplifying the above expression we shall proceed as follows:
<span>x-2/(x^2+4x-12)
factoring out the denominator we get:
</span>x^2+4x-12
=x^2-2x+6x-12
=x(x-2)+6(x-2)
=(x+6)(x-2)
thus our expression will be:
(x-2)/[(x+6)(x-2)]
simplifying the above we get:
1/(x+6)
Answer: 1/(x+6)
9514 1404 393
Answer:
(b) Ingrid’s claim is correct, but her steps are incorrect
Step-by-step explanation:
As nearly as we can tell from the problem statement, Ingrid's table looks like the one attache. We have highlighted the incorrect steps.
Step 3 should read ...
0·cos(y) -(-1)·sin(y) = sin(y) . . . . . . . . . sin(-π/2) = -1
Step 4 should read ...
0 + sin(y) = sin(y)
So, Ingrid's conclusion is correct, but her steps are incorrect.