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A company wants to construct an open rectangular box with a volume of 250 in^3 so that the length of its base is 2 times its width. Express the surface area, S, of the box as a function of the width w.

User MiSimon
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2 Answers

4 votes

Answer: S(w) = 4w^2 + 6wh

Explanation:

The volume of an open rectangular box is given by:

Volume = Length × Width × Height

In this case, the length is 2 times the width, so the length is 2w. Let the height be h.

Given that the volume is 250 in^3, we can write the equation as:

250 = 2w × w × h

Now, we need to express the height (h) in terms of w. Solve for h:

h = 250 / (2w^2)

The surface area of an open rectangular box is given by the formula:

S = 2lw + 2lh + 2wh

Substitute the values:

S = 2(2w)w + 2(2w)h + 2wh

Simplify:

S = 4w^2 + 4wh + 2wh

Simplify further:

S = 4w^2 + 6wh

So, the surface area of the box, S, can be expressed as a function of the width w as:

S(w) = 4w^2 + 6wh

User Viktor Bardakov
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0 votes

Answer:


\sf S(w) = 4w^2 + (750)/(w)

Explanation:

To express the surface area, S, of an open rectangular box as a function of the width, we need to first find the length and height of the box.

Let:

Width of the box = w inches

Length of the box = 2w inches (since the length is 2 times the width)

Height of the box = h inches

The volume of the box (V) is given by:


\sf V = \textsf{length} * \textsf{width} * \textsf{height} \\\\ = (2w) \cdot w \cdot h \\\\ = 2w^2h

We are given that the volume is 250 in^3, so:


\sf 2w^2h = 250

Now, we can express h in terms of w as follows:


\sf h = (250)/(2w^2) \\\\ = (125)/(w^2)

The surface area (S) of the open rectangular box is given by:


\sf S = 2lw + 2lh + 2wh

Substitute the expressions for l and h:


\sf S = 2(2w)(w) + 2(2w)\left((125)/(w^2)\right) + 2w\left((125)/(w^2)\right)

Simplify this expression:


\sf S = 4w^2 + (500)/(w) +( 250)/(w)

Now, we can express S as a function of the width (w):


\sf S(w) = 4w^2 + (750)/(w)

So, the surface area of the open rectangular box is given by:


\sf S(w) = 4w^2 + (750)/(w)

User Dillanm
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