A straight pipe 1 yard in length was marked off in fourths and also in thirds. If the pipe was then cut into separate pieces at
1 answer:
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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:
<h2>y - 5 = -1(x + 3)</h2>Step-by-step explanation:
The point-slope form of an equation of a line:
m - slope
We have m = -1 and the point (-3, 5). Substitute:
Answer:
30
Step-by-step explanation
<h3>In order to find the Least common multiple we write each of the factor only once from each of the expression and for the common expression we take their LCM as maximum of the exponent in all expressions
as here in the question exponent of (x+1) are 3 and 8 so we take exponent 8
likewise for (x-4) we shall take maximum of 2 and 5 which is 5
so our expression for Least common multiple will be
2X3X5 X
30