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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To do solve this you must isolate x. First subtract 16 to both sides (what you do on one side you must do to the other). Since 16 is being added to x, subtraction (the opposite of addition) will cancel it out (make it zero) from the left side and bring it over to the right side.
x + (16 - 16) = 24 - 16
x = 8
Check:
8 + 16 = 24
24 = 24
Hope this helped!
~Just a girl in love with Shawn Mendes
Hello there,
Part 1:
Perimeter of Red = 11+11+11+11
= 44
Perimeter of Blue= 6+6+6+6
= 24
Ratio: Red : Blue
44 : 24(divide by 4)
11 : 6
Part 2:
Area of Red= 11 x 11
= 121
Area of Blue= 6 x 6
= 36
Ratio: Red : Blue
121: 36
Hope this helps :))
~Top
The first blank on the left side should be 0 I think