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Sagot :
Answer:
The maximum volume of the box is:
[tex]V =\frac{5}{3}\sqrt{\frac{5}{3}}[/tex]
Step-by-step explanation:
Given
[tex]Surface\ Area = 10m^2[/tex]
Required
The maximum volume of the box
Let
[tex]a \to base\ dimension[/tex]
[tex]b \to height[/tex]
The surface area of the box is:
[tex]Surface\ Area = 2(a*a + a*b + a*b)[/tex]
[tex]Surface\ Area = 2(a^2 + ab + ab)[/tex]
[tex]Surface\ Area = 2(a^2 + 2ab)[/tex]
So, we have:
[tex]2(a^2 + 2ab) = 10[/tex]
Divide both sides by 2
[tex]a^2 + 2ab = 5[/tex]
Make b the subject
[tex]2ab = 5 -a^2[/tex]
[tex]b = \frac{5 -a^2}{2a}[/tex]
The volume of the box is:
[tex]V = a*a*b[/tex]
[tex]V = a^2b[/tex]
Substitute: [tex]b = \frac{5 -a^2}{2a}[/tex]
[tex]V = a^2*\frac{5 - a^2}{2a}[/tex]
[tex]V = a*\frac{5 - a^2}{2}[/tex]
[tex]V = \frac{5a - a^3}{2}[/tex]
Spit
[tex]V = \frac{5a}{2} - \frac{a^3}{2}[/tex]
Differentiate V with respect to a
[tex]V' = \frac{5}{2} -3 * \frac{a^2}{2}[/tex]
[tex]V' = \frac{5}{2} -\frac{3a^2}{2}[/tex]
Set [tex]V' =0[/tex] to calculate a
[tex]0 = \frac{5}{2} -\frac{3a^2}{2}[/tex]
Collect like terms
[tex]\frac{3a^2}{2} = \frac{5}{2}[/tex]
Multiply both sides by 2
[tex]3a^2= 5[/tex]
Solve for a
[tex]a^2= \frac{5}{3}[/tex]
[tex]a= \sqrt{\frac{5}{3}}[/tex]
Recall that:
[tex]b = \frac{5 -a^2}{2a}[/tex]
[tex]b = \frac{5 -(\sqrt{\frac{5}{3}})^2}{2*\sqrt{\frac{5}{3}}}[/tex]
[tex]b = \frac{5 -\frac{5}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]
[tex]b = \frac{\frac{15 - 5}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]
[tex]b = \frac{\frac{10}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]
[tex]b = \frac{\frac{5}{3}}{\sqrt{\frac{5}{3}}}[/tex]
Apply law of indices
[tex]b = (\frac{5}{3})^{1 - \frac{1}{2}}[/tex]
[tex]b = (\frac{5}{3})^{\frac{1}{2}}[/tex]
[tex]b = \sqrt{\frac{5}{3}}[/tex]
So:
[tex]V = a^2b[/tex]
[tex]V =\sqrt{(\frac{5}{3})^2} * \sqrt{\frac{5}{3}}[/tex]
[tex]V =\frac{5}{3} * \sqrt{\frac{5}{3}}[/tex]
[tex]V =\frac{5}{3}\sqrt{\frac{5}{3}}[/tex]
The maximum volume of the box which has a 10 m² surface area is given below.
[tex]\rm V_{max} = \dfrac{5}{3} *\sqrt{\dfrac{5}{2}}[/tex]
What is differentiation?
The rate of change of a function with respect to the variable is called differentiation. It can be increasing or decreasing.
We want to construct a box with a square base and we currently only have 10 m² of material to use in the construction of the box.
The surface area = 10 m²
Let a be the base length and b be the height of the box.
Surface area = 2(a² + 2ab)
2(a² + 2ab) = 10
a² + 2ab = 5
Then the value of b will be
[tex]\rm b = \dfrac{5-a^2}{2a}[/tex]
The volume of the box is given as
V = a²b
Then we have
[tex]\rm V = \dfrac{5-a^2 }{2a}* a^2\\\\V = \dfrac{5a - a^3}{2}\\\\V = \dfrac{5a}{2} - \dfrac{a^3}{2}[/tex]
Differentiate the equation with respect to a, and put it equal to zero for the volume to be maximum.
[tex]\begin{aligned} \dfrac{dV}{da} &= \dfrac{d}{da} ( \dfrac{5a}{2} - \dfrac{a^3}{2} ) \\\\\dfrac{dV}{da} &= 0 \\\\\dfrac{5}{2} - \dfrac{3a^2 }{2} &= 0\\\\a &= \sqrt{\dfrac{5}{2}} \end{aligned}[/tex]
Then the value of b will be
[tex]b = \dfrac{5-\sqrt{\dfrac{5}{2}} }{2*\sqrt{\dfrac{5}{2}} }\\\\\\b = \sqrt{\dfrac{5}{2}}[/tex]
Then the volume will be
[tex]\rm V = (\sqrt{\dfrac{5}{2}} )^2*\sqrt{\dfrac{5}{2}} \\\\V = \dfrac{5}{3} *\sqrt{\dfrac{5}{2}}[/tex]
More about the differentiation link is given below.
https://brainly.com/question/24062595
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