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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now, you're asked to use it when ln(1.38), which is just another way of saying x = 1.38
so set x = 1.38 and see what "y" is
Answer: 173 months MY GUY!
Step-by-step explanation:
We know that
the distance from the centroid of the triangle to one of the vertices is the radius of the circle <span>required to inscribe an equilateral triangle.
[distance </span>centroid of the triangle to one of the vertices]=(2/3)*h
h=the <span>altitude of the equilateral triangle-----> 5.196 in
so
</span>[distance centroid of the triangle to one of the vertices]=(2/3)*5.196
[distance centroid of the triangle to one of the vertices]=3.464 in----> 3.5 in
the radius is equal to the distance of the centroid of the triangle to one of the vertices
hence
the radius is 3.5 in
the answer is
the radius is 3.5 in
5 -3 * 5 5 simplified = 160!!!