36 cups ........................................
23−11 Enter your answer as a fraction in simplest form by filling in the boxes. 30 30
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
(Image Attached)
Answer:
See attachment
Step-by-step explanation:
The given system is
2x-y=4
x-y=-2
We write the system in slope-intercept form:
y=2x-4
y=x+2
The first equation has slope of 2 and y-intercept at (0,-4)
When y=0, x=2.
We plot (0,-4) and (2,0) to graph this first equation.
The second equation has a slope of 1 and a y-intercept at (0,2)
Also when y=0, x=-2.
We plot (0,2) and (-2,0) and draw a straight line through them.
The graph of the system is shown in the attachment.
