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 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.
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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]
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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]
Learn more
- Find out the area of a cube https://brainly.com/question/12613605
- What is the volume of each prism? https://brainly.com/question/414021
- The volume of a rectangular prism https://brainly.com/question/11613210
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
