The base of a box is a rectangle. The width of the box is half its length. The height of the box is 0.5m. Find the volume of the box if the area of the base is 1.08m² less than the combined area of the sides.

Let x be the length and y be the width.
We know already
[tex]y= \frac{x}{2} [/tex]
Area of the base [tex]xy[/tex]
Area of the sides: [tex](x+y)0.25[/tex]
We get the equation [tex]xy=0.25(x+y)-1.08[/tex]
We will solve the system:
[tex]y= \frac{x}{2}\\xy=0.5(x+y)-1.08[/tex]
The above system has no real solution, please check your values.

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Аноним Аноним · Answer 2 · 2018-11-29T08:00:33+01:00

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

We assume the box is open-top.

If the width of the box is represented by w, then its length is 2w. The area of the base is then w·2w = 2w².

The combined area of the sides is the product of the perimeter of the base, 2(w+2w) = 6w and the height, 0.5, so is 3w.

Since the base area is 1.08 m² less than the lateral area, we have

... 2w² = 3w -1.08

... 2w² -3w +1.08 = 0 . . . . rearrange to standard form

... w² -1.5w +0.54 = 0 . . . make the leading coefficient 1 (divide by 2)

... (w -0.9)(w -0.6) = 0 . . . factor

The width of the box may be either 0.6 meters or 0.9 meters.

The volume is the product of width, length, and height

V = w·2w·0.5 = w²

V = 0.36 m³ . . . or . . . 0.81 m³. . . . . . (there are two possible answers)