1 answer

A box with a square base .6 4. Compute x + 3x4 + 2x3 + 1...

Question:

a box with a square base
.6 4. Compute x + 3x4 + 2x3 + 1 -da. 24 일 5. Let F(x) = tet-2+tº +1 dt, find F(2). tt +3 0
-. A box with a square base and open top must have a volume of 500 cm. Find the dimensions of the box which minimize the amou
.6 4. Compute x + 3x4 + 2x3 + 1 -da. 24 일 5. Let F(x) = tet-2+tº +1 dt, find F'(2). tt +3 0
-. A box with a square base and open top must have a volume of 500 cm. Find the dimensions of the box which minimize the amount of material to be used. 2. Draw the graph of f(x) = x ln(1x) - (x - 4) In(x - 41).

Answers

Let x be the length,

Let y be the height,

Let z be the width.

The volume of the box is given by,

V = xyz

for square base x = z,

Ꮖ Ꮖ ;

V = x^{2}y

500 = x^{2}y

y = \frac{500}{x^{2}}

The surface area of the open-top box is given by,

S = xz + 2xy+2zy

put x = z,

S = x(x) + 2xy+2xy

S = x^{2} + 4xy

plugin equation for y,

,S = x^{2} + 4x\left ( \frac{500}{x^{2}} \right )

S = x^{2} + \frac{2000}{x}

diff wrt x,

\frac{\mathrm{d} S}{\mathrm{d} x} = 2x - \frac{2000}{x^{2}}

For minimum surface are,

\frac{\mathrm{d} S}{\mathrm{d} x} =0

2x - \frac{2000}{x^{2}} =0

x - \frac{1000}{x^{2}} =0

x^{3} = 1000

x = 10

x = z= 10

Now find the y value,

y = \frac{500}{10^{2}} = 5

Hence the dimensions of the box are,

x = z= 10

y = 5

I hope this answer helps,
Thanks,
Keep posting questions.

.

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