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:
The answer is A: 5x + 10y > 30. That is, the combination of boxes must be greater than 30 since the requirement is to have more than 30 lb of nails.
Step-by-step explanation:
The worker can buy a combination of boxes, as long as the total is greater than 30 lb. Multiply 5 lb by x and add that to 10 lb times y to get the total, which must exceed 30 lb.
Answer:
cos (105°) = - 2.6
csc (105°) = 1.03
Step-by-step explanation:
Given that cos (-105°) = - 0.26 {The negative sign is due to the angle - 105° lies in the third quadrant where cos value is negative}
Again, given that csc (- 105°) = - 1.03 {{The negative sign is due to the angle - 105° lies in the third quadrant where csc value is negative}
Now, cos (105°) = - 2.6, because 105° lies in the second quadrant and here cos value is negative.
And csc (105°) = 1.03, because 105° lies in the second quadrant and here csc value is positive. (Answer)
Answer:
Step-by-step explanation:
32
7a
TSA = 2lw<span> + 2</span>lh<span> + 2</span><span>wh
TSA = 140 + 70 + 100 = 310 ft^2
7b
310 x $4 =$1240
cost = </span>$1240
7c
V= wlh = <span>7×10×5 = </span><span>350 ft^3</span>