PLEASE HELP ! Find the zeros of the given function.
2 answers:
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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)
Hello!
Remember that the symbols: ≤ and ≥ are graphed as a solid line. While the symbols: < and > are graphed as a dotted line.
Also, before graphing, it would be better to convert both equations to slope-intercept form.
y ≤ x + 1 is already in slope-intercept form.
y + x ≤ -1 is not written in slope-intercept form. (Slope-intercept form: y = mx + b)
y + x ≤ - 1 (subtract x from both sides)
y ≤ -x - 1
Graphing those lines, you get the graph below. You can see that Part C best represents the solution set systems of inequalities, because that is where both of the shaded lines intersect.
Answer: Part C
