The dimensions of a rectangular prism can be expressed as width w, length w+4, and height 5-w. What is the approximate width (w) that will maximize the volume of the tank?

The approximate value of w that maximize the volume

V = -w³ + w² + 20w

is w = 3 units

Step-by-step explanation:

Volume of rectangular prism is given by the formula:

V = Bh

B is the base, and

B = lw

Where l is the length of the rectangle, and w is the width.

B = w(w + 4)

= w² + 4w

V = Bh

= (w² + 4w)(5 - w)

= 5w² - w³ + 20w - 4w²

V = -w³ + w² + 20w

To obtain the approximate value of w that maximizes the volume, let us solve:

-w³ + w² + 20w = 0

w(-w² + w + 20) = 0

w = 0

Or

-w² + w + 20 = 0

Using the quadratic formula,

w = [-b±√(b² - 4ac)]/2a

Where a = -1, b = 1, and c = 20.

w = {-1 ±√[(1 - 4(1)(20)]}/2(-1)

= (-1 ±√25)/(-2)

= 1/2 ± -5/2

w = 1/2 + 5/2

= -4/2 = -2

w = 1/2 - 5/2

= 6/2 = 3

Now we have the following values for w.

w = 0 or -2 or 3

The maximum of these is 3, and it is what we are looking for.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)