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:
0.800
Step-by-step explanation:
Answer:
13
Step-by-step explanation:
using BODMAS
13-11=2
SO 2*2=4
4/4=1
1*72=72
THEREFORE it is 85-72 =13
Given : Kishore bought a basket containing 30 oranges for rs 310. If he sells the oranges at rs 12 each , what Would be it's profit or loss?
Solution :
Cost price of oranges = rs.310
As we know that each oranges cost rs.12
So, selling price of oranges is 30 × 12 = rs.360
Now, <u>finding profit or loss</u>
Profit = Selling Price - Cost Price
Profit = rs.360 - rs.310
Profit = rs.50