The ages are consecutive even numbers.
The youngest child is (x+3) years old.
Every even number is 2 more than the one before it.
So the three ages are
(x + 3), (x + 5), and (x + 7).
Their sum is x+3 + x+5 + x+7 = 3x + 15
But we're told that their sum is 42, so we can write
3x + 15 = 42 <== the equation
Subtract 15 from each side: 3x = 27
Divide each side by 3 : x = 9 <== the solution
So the three ages are 12, 14, and 16.
The oldest child is 16 years old.
Answer: D 87%
Step-by-step explanation:
Ap-ex
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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