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Sphinxa [80]
3 years ago
15

If 600 cm2 of material is available to make a box with a square base and a closed top, find the maximum volume of the box in cub

ic centimeters. Answer to the nearest cubic centimeter without commas. For example, if the answer is 2,000 write 2000.
Mathematics
2 answers:
natali 33 [55]3 years ago
8 0
<h2>Answer:</h2>

The maximum volume of the box is:

              1000 cm³

<h2>Step-by-step explanation:</h2>

Let x be the length of the square base

and h be the height(h) of the box.

As we know that the length(l) and width(w) of the box is same( since the base is in the shape of square)

As we know that the surface area of box is given by:

Surface\ Area=2(lw+wh+hl)\\\\\\i.e.\\\\\\Surface\ Area=2(x^2+xh+xh)\\\\\\Surface\ Area=2(x^2+2hx)

We are given surface area of box=600 cm²

Hence,

2(x^2+2hx)=600\\\\i.e.\\\\x^2+2xh=300\\\\i.e.\\\\h=\dfrac{150}{x}-\dfrac{x}{2}

The volume of box is given by:

Volume(V)=l\times w\times h\\\\\\Volume=x^2h\\\\\\Volume=x^2(\dfrac{150}{x}-\dfrac{x}{2})

Hence,

V=150x-\dfrac{x^3}{2}

Now, for maxima or minima we have derivative equal to zero.

i.e.\\\\\\\dfrac{dV}{dx}=0\\\\\\i.e.\\\\\dfrac{d}{dx}(150x-\dfrac{x^3}{2})=0\\\\\\i.e.\\\\\\150-\dfrac{3x^2}{2}=0\\\\\\i.e.\\\\\\150=\dfrac{3}{2}x^2\\\\\\x^2=100\\\\\\i.e.\\\\\\x=\pm 10

Now,

\dfrac{d^2V}{dx^2}=\dfrac{-6x}{2}\\\\\\=-3x

Now, as we know if for a given x

\dfrac{d^2V}{dx^2}

then that x is a point of maxima.

Hence, when we put x=10 we get:  \dfrac{d^2V}{dx^2}

Hence, x=10 is a point of maxima

Also, the value of V at x=10 is:

V=150\times 10-\dfrac{(10)^3}{2}\\\\\\V=1500-\dfrac{1000}{2}\\\\\\V=1500-500\\\\V=1000\ cubic\ cm.

       Hence, the maximum volume of the box is:

                     1000 cm³

juin [17]3 years ago
6 0
Let the sides of square base be "x" and height be "h" first come up with expressions for surface area and volume<span>SA=2<span>x2</span>+4hx=600</span> <span>V=h<span>x2</span></span>Next get h in terms of x and sub it into Volume equation<span>h=<span><span>300−<span>x2</span></span><span>2x</span></span></span> <span>→V=(<span><span>300−<span>x2</span></span><span>2x</span></span>)<span>x2</span></span> <span>V=150x−<span>12</span><span>x3</span></span>maximize volume by setting derivative equal to 0<span><span><span>dV</span><span>dx</span></span>=150−<span>32</span><span>x2</span>=0</span>solve for x<span>x=10</span><span> which means h =10 max volume occurs when box is a cube 10^3 = 1000 cm^3</span>
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