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Afina-wow [57]
3 years ago
7

The polynomial -x^3-x^2+12x represents the volume of a rectangular aquatic tank in Cubic feet. The length of the tank is (x+4).

A. Use synthetic formula to help factor the volume of the polynomial. How many linear factors should you look for? B. What are the dimensions of the tank? C.Find the value of x that will maximize the volume of the box. D. What is the maximum volume?

Mathematics
1 answer:
Sliva [168]3 years ago
6 0

Answer:

  A. x(x +4)(3 -x) . . . . a cubic will have 3 linear factors

  B. The dimensions are x, x+4, 3 -x, or 1.694 ft, 5.694 ft, 1.306 ft

  C. x = (√37 -1)/3 ≈ 1.694254 ft

  D. Maximum volume: ≈ 12.597 ft³

Step-by-step explanation:

A. The first attachment shows the synthetic division of the given polynomial by x+4. The result is -x^2 +3x = x(3 -x). Hence the factored polynomial is ...

  x(3-x)(x+4) . . . . . . shown as positive

Since the polynomial is a cubic, we know there will be 3 roots. For this problem, that means we expect 3 factors.

__

B. We are apparently to assume that the dimensions of the tank correspond to the three factors:

  • x
  • 3-x
  • x+4

For the maximum size tank (see part C), the corresponding measurements in feet are

  • 1.694 ft
  • 1.306 ft
  • 5.694 ft

__

C. Tank volume will be maximized when the derivative of volume with respect to x is zero. The derivative polynomial is ...

  dV/dx = -3x^2 -2x +12

Solving this by the usual methods, we find the positive value of x to be ...

  x = (-1 +√37)/3 ≈ 1.694254 . . . . feet

__

D. Evaluating the polynomial for this value of x gives ...

  volume ≈ 12.5972 . . . cubic feet

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\huge \boxed{\mathbb{QUESTION} \downarrow}

  • Solve 2m + n = 2 and 3m - 2n = 3 using substitution.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

We can use the substitution method to solve linear equations of this form. Let's solve for m & n.

\left. \begin{array}  { l  }  { 2 m + n = 2 } \\ { 3 m - 2 n = 3 } \end{array} \right.

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

2m+n=2, \: 3m-2n=3

Choose one of the equations and solve it for m by isolating m on the left-hand side of the equal sign.

2m+n=2

Subtract n from both sides of the equation.

2m=-n+2

Divide both sides by 2.

m=\frac{1}{2}\left(-n+2\right)  \\

Multiply 1/2 times -n+2.

m=-\frac{1}{2}n+1  \\

Substitute -\frac{n}{2}+1\\ for m in the other equation, 3m-2n=3.

3\left(-\frac{1}{2}n+1\right)-2n=3  \\

Multiply 3 times -\frac{n}{2}+1\\.

-\frac{3}{2}n+3-2n=3  \\

Add -\frac{3n}{2}\\ to -2n.

-\frac{7}{2}n+3=3  \\

Subtract 3 from both sides of the equation.

-\frac{7}{2}n=0  \\

Divide both sides of the equation by -\frac{7}{2}\\, which is the same as multiplying both sides by the reciprocal of the fraction.

\large \underline{\underline{ \bf \: n=0 }}

Substitute 0 for n in m=-\frac{1}{2}n+1\\. Because the resulting equation contains only one variable, you can solve for m directly.

\large \underline{ \underline{\bf \: m=1 }}

The system is now solved.

\huge \boxed{ \boxed{ \bf \: m=1, \: n=0 }}

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