Answer:
legs limit the number of chairs that can be built
Step-by-step explanation:
The maximum number of chairs that can be built will be the minimum of the number of parts divided by the number of parts needed for each chair, as computed across the different kinds of parts required.
seats: 12 available, used 1 per chair: 12/1 = 12 chairs possible
backs: 15 available, used 1 per chair: 15/1 = 15 chairs possible
legs: 44 available, used 4 per chair: 44/4 = 11 chairs possible
The maximum number of chairs that can be built will be the minimum of 12, 15, and 11. That is, 11 chairs can be built, limited by the number of available legs.
Answer:
0.007%
Step-by-step explanation:
4000/28=0.007%
While it's pretty obvious to most of us that
-13x=90-2y
-6x=48-2y
is a system of linear equations, it'd be well to include that info plus the instructions "solve this system of linear equations."
Subtract the 2nd equation from the first:
-13x=90-2y
+6x=-48+2y
-----------------------
-7x = 42. Then x = -42/7, or x = 6.
Now subst. 6 for x in either one of the given equations. Suppose we use the 2nd equation:
-6x=48-2y
Then -6(6)=48-2y, or -36 = 48 - 2y, or 2y = 48+ 36 = 84. Then y = 42.
The solution is (6, 42).
Given the numbers on the computer 123 is shown very complicated. Well it’s not given all the numbers and mind thinking if u add formula 1 that shows that I have no idea what to tell you and ur stuck
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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