Question

Can I please get help assignment due in an hour.

Answer 1

A graphing calculator is recommended.

A rectangular box with a volume of
\[2\sqrt{3} ft^3\]
has a square base as shown below. The diagonal of the box (between a pair of opposite corners) is 1 ft longer than each side of the base.

(a) If the base has sides of length x feet, show that
\[x^6 − 2x^5 − x^4 + 12 = 0\]
Let x be the length, in ft, of each side of the base and let h be the height. The length of the diagonal on the base is

\[\sqrt{x^2 + }= \sqrt{?} \]
and hence the length of the diagonal of the box is

\[\sqrt{?} = x + 1\]
This gives us
h=

Answer 2

.
So, the volume of the box is
\[V = 2\sqrt{3} = hx^2\]
=\[=x^2( )\]



Therefore,
\[2\sqrt{3} = x^2( )\]
\[ 12 = x^4 ( )\]
\[? = 0\]
(b) Show that two different boxes satisfy the given conditions. Find the dimensions in each case, correct to the nearest hundredth of a foot.

Solving the equation from part (a), we have the following (Enter your answers as a comma-separated list.)
x=

For the larger value of x,
width = length ft
height ft.

For the smaller value of x,
width = length ft
height ft.

Answer 3

Just the Base: \(x^{2} + x^{2} = (x+1)^{2}\)

Volume: \(h\cdot x^{2} = 2\sqrt{3}\)

We'll need to solve the "Base" Equation if we've any hope of finding the height, h.

Answer 4

I am confused :/

Answer 5

Base is a square x by x.
Therefore the diagonal of the base (NOT the diagonal of the box) = sqrt(x^2 + x^2) = (sqrt(2))x
Height is h.
So the square of the diagonal of the box = (h^2 + 2x^2)
The box diagonal is one foor longer than the side = (x + 1). So square it and equate to the last line.
h^2 + 2x^2 = (x + 1)^2 ----- equation (1)

Volume of box = x^2h = 2sqrt(3)
Square both sides
x^4h^2 = 4(3) = 12
h^2 = 12/x^4
Put this in equation (1)

Answer 6

Put h^2 = 12/x^4
in h^2 + 2x^2 = (x + 1)^2

12/x^4 + 2x^2 = x^2 + 2x + 1
Multiply by x^4
12 + 2x^6 = x^6 + 2x^5 + x^4
x^6 - 2x^5 - x^4 + 12 = 0
That is your part a)

Answer 7

Any questions?

Answer 8

Part b)

Using graphing calculator equation a) has two roots:
x = 2.23, 1.67

When x = 2.23 ft, h = 0.7 ft

When x = 1.67 ft, h = 1.24 ft.

Answer 9

Thanks a lot that helped me so much!

Answer 10

you are welcome.

Answer 11

The part b has an incorrect part. When x=1.67, 1.24ft seems to be incorrect.

Answer 12

Diagonal of the BOX. Missed that. I was thinking diagonal of the BASE. Good call, @ranga !

Answer 13

Volume = x^2h = 2sqrt(3) = 3.4641
h = 3.4641 / x^2
x = 2.23, h = 3.4641 / 2.23^2 = 0.6965 = 0.70 (rounded to the nearest hundredth of a foot)
x = 1.67, h = 3.4641 / 1.67^2 = 1.2421 = 1.24 (rounded to the nearest hundredth of a foot)


So I am getting:
x = 2.23, h = 0.70 and
x = 1.67, h = 1.24

They have asked us to specifically round x and h off to the nearest hundredth. Therefore, if you try to find the volume with x^2h in the two cases above it won't exactly come to 2sqrt(3) because the rounding off error gets magnified when x and x and h are multiplied.

Instead of x = 1.67, had I kept it to three decimal places, then x = 1.667 and
h = 3.4641 / 1.667^2 = 1.247 and when rounded to the nearest hundredth it will be 1.25 instead of 1.24. But because they asked us to round x to the second decimal I had to use 1.67 for x and I got 1.24 for h.


Thanks @tkhunny