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Snowcat [4.5K]
3 years ago
6

A rectangular storage container with an open top is to have a volume of 24 cubic meters. The length of its base is twice the wid

th. Material for the base costs 13 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.
Mathematics
1 answer:
Mariana [72]3 years ago
4 0

Answer:

419.25

Step-by-step explanation:

The calculation of the cost of materials for the cheapest such container is shown below:-

We assume

Width = x

Length = 2x

Height = h

where, length = 2 \times width

Base area = lb

= 2x^2

Side area = 2lh + 2bh

= 2(2x)h + 2(x)h

= 4xh + 2xh

Volume = 24 which is lbh = 24

h = \frac{24}{2x^2} \\\\ h = \frac{12}{x^2}

Now, cost is

= 13(2x^2) + 9(4xh + 2xh)\\\\ = 13(2x^2) + 9(4x + 2x)\times \frac{12}{x^2} \\\\ = 26x^2 + \frac{648}{x}

now we have to minimize C(x)

So, we need to compute the C'(x)

= 52x - \frac{648}{x^2}

C"(x)  = 52x - \frac{1,296}{x^3}

now for the critical points, we will solve the equation C'(x) = 0

= 52x - \frac{648}{x^2} = 0\\\\ x = \frac{648}{52}^{\frac{1}{3}}

C" = ((\frac{648}{52} ^{\frac{1}{3} } = 52 + \frac{1296}{(\frac{648}{52})^\frac{1}{3} )^3}\\\\ = 52 + \frac{1296}{\frac{648}{52} } >0

So, x is a point of minima that is

= (\frac{648}{52} )^\frac{1}{3}

Now, Base material cost is

= 13(2x^2)\\\\ = 26(\frac{648}{52} )^\frac{2}{3}

= 139.75

Side material cost is

= \frac{648}{x} \\\\ = \frac{648}{(\frac{648}{52})^\frac{1}{3}  }

= 279.50

and finally

Total cost is

= 139.75 + 279.50

= 419.25

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A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

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If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

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