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
To learn more about Maximization visit
brainly.com/question/14682292
#SPJ4
Complete Question
(Image Attached)
Recall that the slope is the "average rate of change" of any function.

5 is your missing factor because 9*5=45 btw dont use x for multiplication use this *
Answer:
They did not distribute,
8-3(2x-5)
8-6x+15
23-6x
Step-by-step explanation:
8-3(2x-5)
8-6x+15
23-6x
Have you considered applying the Law of Cosines?
For angle B: b^2 = a^2 + c^2 - 2ac*cos B
a, b and c are all known, leaving cos B as the only unknown. Find cos B, and then find angle B itself. From the diagram we know that angles A, B and C are all acute.