An open top box with a square base has a surface area of 100 square inches. express the volume of the box as a function of the length of the edge of the base. what is its domain?

Respuesta :

Let the base be a square of side length = a ft, and height h ft.

The box has a base of area: [tex]a \cdot a = a^2 (square ft) [/tex], 

and 4 lateral faces of area ah (square ft).


thus the total area of the box is [tex]a^2+4ah[/tex].

we can reduce the number of variables as follows:

[tex]a^2+4ah=100\\\\4ah=100-a^2\\\\h= \frac{100-a^2}{4a}= \frac{25}{a}- \frac{a}{4} [/tex]



The volume of the box is given by the function:

[tex]V(a)=a \cdot a \cdot (\frac{25}{a}- \frac{a}{4} )=25a- \frac{a^3}{4} [/tex].



To find the domain we solve 

[tex]25a- \frac{a^3}{4} \ \textgreater \ 0\\\\a(25- ( \frac{a}{2} )^{2} )\ \textgreater \ 0\\\\a(5- \frac{a}{2})(5+ \frac{a}{2} )\ \textgreater \ 0[/tex]


the roots of the expression are 0, 10 and -10.    Since the overall sign of the expression is negative, we have the following sign table:

++++++++++++(-10)----------(0)++++++++++(10)--------------------

thus the solution of the inequation is (-infinity, -10) union (0, 10)

but we have another condition: a>0, since each side must be positive.


combining these 2 conditions, we have the domain: (0, 10)


Answer:  [tex]V(a)=25a- \frac{a^3}{4}[/tex], Domain (0, 10)



  • The volume = 25x - ¹/₄x³ (in cubic inches)
  • Its domain (0, 10)

Further explanation

Given:

An open-top box with a square base has a surface area of 100 square inches.

Question:

  • Express the volume of the box as a function of the length of the edge of the base.
  • What is its domain?

The Process:

  • Let the length of the edge of the base = x
  • Let height = h

Part-1: The surface area

Let us arrange the equation to get the surface area of ​​the box with a square base. Recall that the box is without a lid and its surface area is 100 square inches.

[tex]\boxed{ \ Surface \ area = area \ of \ base + (4 \times area \ of \ rectangle) \ }[/tex]

[tex]\boxed{ \ x^2 + (4 \cdot x \cdot h) = 100 \ }[/tex]

[tex]\boxed{ \ x^2 + 4xh = 100 \ }[/tex]

From the above equation, we set it again so that "xh" is the subject on the left.

Both sides are subtracted by x².

[tex]\boxed{ \ 4xh = 100 - x^2 \ }[/tex]

Both sides are divided by 4.

[tex]\boxed{ \ xh = \frac{100 - x^2}{4} \ }[/tex] ... (Equation-1)

That is the strategy we have prepared.

- - - - - - - - - -

Part-2: The volume

[tex]\boxed{ \ Volume \ of \ the \ box = length \times width \times height \ }[/tex]

[tex]\boxed{ \ Volume = x \cdot x \cdot h \ }[/tex]

[tex]\boxed{ \ Volume = x^2h \ }[/tex] ...(Equation-2)

Substitution Equation-1 into Equation-2.

[tex]\boxed{ \ Volume = x(xh) \ }[/tex]

[tex]\boxed{ \ Volume = x \bigg( \frac{100 - x^2}{4} \bigg) \ }[/tex]

[tex]\boxed{ \ Volume = \frac{100x - x^3}{4} \ }[/tex]

[tex]\boxed{ \ Volume = 25x - \frac{1}{4}x^3 \ }[/tex]

Thus, an expression of the volume of the box as a function of the length of the edge of the base is [tex]\boxed{\boxed{ \ Volume = 25x - \frac{1}{4}x^3 \ }}[/tex]

- - - - - - - - - -

Part-3: The domain of volume

The value of volume must always be positive, i.e., V > 0.

[tex]\boxed{ \ 25x - \frac{1}{4}x^3 > 0 \ }[/tex]

Both sides are multiplied by 4.

[tex]\boxed{ \ 100x - x^3 > 0 \ }[/tex]

Both sides are multiplied by -1, notice the change in the sign of the inequality.

[tex]\boxed{ \ x^3 - 100x < 0 \ }[/tex]

[tex]\boxed{ \ x(x^2 - 100) < 0 \ }[/tex]

[tex]\boxed{ \ x(x - 10)(x + 10) < 0 \ }[/tex]

We get [tex]\boxed{ \ x = 0, \ x = 10, \ and \ x = - 10 \ }[/tex].

  • Since the values of x cannot be negative, x = -10 are promptly rejected.
  • For x = 0 can be used as one of the domain limits.

Consider the test of signs:

x(x - 10) (x + 10) is negative to the left of x = 10, and positive to the right of x = 10 on the number line.

Examples of tests:

  • [tex]\boxed{ \ for \ x = 2 \rightarrow 2(2 - 10)(2 + 10) < 0 \ }[/tex]
  • [tex]\boxed{ \ for \ x = 11 \rightarrow 11(11 - 10)(11 + 10) > 0 \ }[/tex]

Remember this form above, [tex]\boxed{\ x(x - 10)(x + 10) < 0 \ }[/tex], the value of the test result must be negative (because < 0).

Thus, the domain of the volume is [tex]\boxed{ \ 0 < x < 10 \ }[/tex] or [tex]\boxed{ \ (0, 10) \ }[/tex]

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Keywords: an open-top box, with, a square base, has a surface area, 100 square inches, express, the volume, as a function, the length, edge, base, what, its domain, test of signs, substitution

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