Yes, it is shown horizontal for both lines and they will never touch, therefore parallel.
Well you could get 7 right
Answer:
The answer is "140".
Step-by-step explanation:
Calculating the average cost:
To minimize average cost we have to find x for wich A'(x)=0
<span>Width = 4.56 inches
Length = 18.26 inches
Height = 7.30 inches
The box has 6 sides that needs to be covered, so let's create an equation to express the total surface area of the box.
A = 2*w*l + 2*w*h + 2*h*l
Now since we know that the length is 4 times the width, let's substitute 4w for l in the above equation and simplify.
A = 2*w*4w + 2*w*h + 2*h*4w
A = 8*w^2 + 2*w*h + 8*h*w
A = 8*w^2 + 2*h*w + 8*h*w
A = 8*w^2 + 10*h*w
A = 2w(4w + 5h)
And now since the area has to be 500, let's express h in terms of W. Substitute 500 for A and then solve for h.
A = 2w(4w + 5h)
500 = 2w(4w + 5h)
250/w = 4w + 5h
250/w - 4w = 5h
(250/w - 4w)/5 = h
50/w - (4/5)w = h
So we now can calculate l (length) and h (height) of the box in terms of w. The equation for the volume is:
V = lwh
Let's substitute 4w for l
V = lwh
V = 4wwh
V = 4hww
V = 4hw^2
And substitute (50/w - (4/5)w) for h
V = 4hw^2
V = 4(50/w - (4/5)w)w^2
V = (200/w - (16/5)w)w^2
V = 200w - (16/5)w^3
Now since we're looking for the largest possible volume, that should bring to mind "first derivative". We can use the power rule to calculate that easily.
V = 200w - (16/5)w^3
V' = 200 - 3(16/5)w^2
V' = 200 - (48/5)w^2
Now the minimum and maximum values of V can only happen where V' equal 0. So let's set it to 0 and calculate w.
V' = 200 - (48/5)w^2
0 = 200 - (48/5)w^2
(5/48)*(48/5)w^2 = (5/48)*200
w^2 = 1000/48 = 125/6
w = (5/6)sqrt(30), approximately 4.564354646
Now let's calculate length and height.
l = (5/6)sqrt(30) * 4 = 18.25741858
h = 50/w - (4/5)w = 50/((5/6)sqrt(30)) - (4/5)((5/6)sqrt(30)) = 7.302967433
Rounding to 2 decimal places gives w=4.56, l=18.26, h=7.30</span>
Answer:
(2x + 3y)(2x - 3y)
Step-by-step explanation:
(2x + 3y)(2x - 3y)
