Answer
Verified
384k+ views
Hint: Let us assume an open box, with a square base with each edge of length a inch, and height of the box h inch.
Let the surface area of the box (not including the top the top) be s, then we have
$s=\left( a\times a \right)+4\left( a\times h \right)$
Thus, the surface area of the open box is $s={{a}^{2}}+4ah$.
We can also write the above equation as
$4ah=s-{{a}^{2}}$
And thus, we have the height of the open box as
$h=\dfrac{s-{{a}^{2}}}{4a}...\left( i \right)$
Complete step-by-step solution:
From the figure, we can clearly say that the volume of this open box is
$V=a\times a\times h$
Using the value of height from equation (i), we get
$V={{a}^{2}}\left( \dfrac{s-{{a}^{2}}}{4a} \right)$
We can simplify this equation by cancelling terms to get,
$V=\left( \dfrac{as-{{a}^{3}}}{4} \right)...\left( ii \right)$
Here, we know that the value of s is constant, and we need to find the value of a for which the volume is maximum.
So, we know that we need to differentiate the volume with respect to a. Thus, we have
$\dfrac{dV}{da}=\dfrac{d}{da}\left( \dfrac{as-{{a}^{3}}}{4} \right)$
So, on differentiation, we get
$\dfrac{dV}{da}=\dfrac{d}{da}\left( \dfrac{as}{4} \right)-\dfrac{d}{da}\left( \dfrac{{{a}^{3}}}{4} \right)$
Hence, we now have the following equation
$\dfrac{dV}{da}=\dfrac{s}{4}-\dfrac{3{{a}^{2}}}{4}$
We now need to equate this differentiation with zero. So, we get
$\dfrac{s}{4}-\dfrac{3{{a}^{2}}}{4}=0$
Solving for a, we get
$3{{a}^{2}}=s$
Or, we can write
$a=\sqrt{\dfrac{s}{3}}$
We still need to see whether this is a maxima point or minima point.
So, ${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}=\dfrac{d}{da}{{\left[ \dfrac{s}{4}-\dfrac{3{{a}^{2}}}{4} \right]}_{a=\sqrt{\dfrac{s}{3}}}}$
On solving this equation, we get
${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}={{\left[ 0-\dfrac{3a}{2} \right]}_{a=\sqrt{\dfrac{s}{3}}}}$
On putting the value of a, we get
${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}=\left[ -\dfrac{3\sqrt{\dfrac{s}{3}}}{2} \right]$
Or, we can write
${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}=\left[ -\dfrac{\sqrt{3s}}{2} \right]$
We can clearly see that ${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}<0$ .
Thus, the volume is maximum at $a=\sqrt{\dfrac{s}{3}}$ .
We know that the surface area is 27 $\text{i}{{\text{n}}^{2}}$ as is given in the question. So, we now get the value
$a=\sqrt{\dfrac{27}{3}}$
And since the value of a can never be negative, the value of a will be
$a=3$ .
Putting the values of a and s in equation (ii), we get
$V=\left( \dfrac{\left( 3\times 27 \right)-{{\left( 3 \right)}^{3}}}{4} \right)$
Thus, we get
$V=\left( \dfrac{\left( 3\times 27 \right)-27}{4} \right)$
On solving the above equation, we get
$V=\left( \dfrac{27}{2} \right)\text{i}{{\text{n}}^{3}}$
Hence, $V=13.5\text{ i}{{\text{n}}^{3}}$
Thus, the maximum volume of the open box is $13.5\text{ i}{{\text{n}}^{3}}$.
Note: We must remember that the surface area of open box with a square base is $s={{a}^{2}}+4ah$ and the surface area of a closed box with a square base is $s=2{{a}^{2}}+4ah$. Here, $a=\sqrt{\dfrac{s}{3}}$ , and since a can not be a negative value, we can say that $\sqrt{s}$ is also positive and thus, $\left[ -\dfrac{\sqrt{3s}}{2} \right]$ will be negative.
Let the surface area of the box (not including the top the top) be s, then we have
$s=\left( a\times a \right)+4\left( a\times h \right)$
Thus, the surface area of the open box is $s={{a}^{2}}+4ah$.
We can also write the above equation as
$4ah=s-{{a}^{2}}$
And thus, we have the height of the open box as
$h=\dfrac{s-{{a}^{2}}}{4a}...\left( i \right)$
Complete step-by-step solution:
From the figure, we can clearly say that the volume of this open box is
$V=a\times a\times h$
Using the value of height from equation (i), we get
$V={{a}^{2}}\left( \dfrac{s-{{a}^{2}}}{4a} \right)$
We can simplify this equation by cancelling terms to get,
$V=\left( \dfrac{as-{{a}^{3}}}{4} \right)...\left( ii \right)$
Here, we know that the value of s is constant, and we need to find the value of a for which the volume is maximum.
So, we know that we need to differentiate the volume with respect to a. Thus, we have
$\dfrac{dV}{da}=\dfrac{d}{da}\left( \dfrac{as-{{a}^{3}}}{4} \right)$
So, on differentiation, we get
$\dfrac{dV}{da}=\dfrac{d}{da}\left( \dfrac{as}{4} \right)-\dfrac{d}{da}\left( \dfrac{{{a}^{3}}}{4} \right)$
Hence, we now have the following equation
$\dfrac{dV}{da}=\dfrac{s}{4}-\dfrac{3{{a}^{2}}}{4}$
We now need to equate this differentiation with zero. So, we get
$\dfrac{s}{4}-\dfrac{3{{a}^{2}}}{4}=0$
Solving for a, we get
$3{{a}^{2}}=s$
Or, we can write
$a=\sqrt{\dfrac{s}{3}}$
We still need to see whether this is a maxima point or minima point.
So, ${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}=\dfrac{d}{da}{{\left[ \dfrac{s}{4}-\dfrac{3{{a}^{2}}}{4} \right]}_{a=\sqrt{\dfrac{s}{3}}}}$
On solving this equation, we get
${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}={{\left[ 0-\dfrac{3a}{2} \right]}_{a=\sqrt{\dfrac{s}{3}}}}$
On putting the value of a, we get
${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}=\left[ -\dfrac{3\sqrt{\dfrac{s}{3}}}{2} \right]$
Or, we can write
${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}=\left[ -\dfrac{\sqrt{3s}}{2} \right]$
We can clearly see that ${{\left. \dfrac{{{d}^{2}}V}{d{{a}^{2}}} \right|}_{a=\sqrt{\dfrac{s}{3}}}}<0$ .
Thus, the volume is maximum at $a=\sqrt{\dfrac{s}{3}}$ .
We know that the surface area is 27 $\text{i}{{\text{n}}^{2}}$ as is given in the question. So, we now get the value
$a=\sqrt{\dfrac{27}{3}}$
And since the value of a can never be negative, the value of a will be
$a=3$ .
Putting the values of a and s in equation (ii), we get
$V=\left( \dfrac{\left( 3\times 27 \right)-{{\left( 3 \right)}^{3}}}{4} \right)$
Thus, we get
$V=\left( \dfrac{\left( 3\times 27 \right)-27}{4} \right)$
On solving the above equation, we get
$V=\left( \dfrac{27}{2} \right)\text{i}{{\text{n}}^{3}}$
Hence, $V=13.5\text{ i}{{\text{n}}^{3}}$
Thus, the maximum volume of the open box is $13.5\text{ i}{{\text{n}}^{3}}$.
Note: We must remember that the surface area of open box with a square base is $s={{a}^{2}}+4ah$ and the surface area of a closed box with a square base is $s=2{{a}^{2}}+4ah$. Here, $a=\sqrt{\dfrac{s}{3}}$ , and since a can not be a negative value, we can say that $\sqrt{s}$ is also positive and thus, $\left[ -\dfrac{\sqrt{3s}}{2} \right]$ will be negative.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Which are the Top 10 Largest Countries of the World?
One cusec is equal to how many liters class 8 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
The mountain range which stretches from Gujarat in class 10 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths