Understand Maxima and Minima with an example. click on the link to Watch the VIDEO explanation: Watch Video
Problem on Maxima and Minima
A sheet of cardboard 8 cm by 5 cm is made into a box by cutting equal sized squares from each corner and folding the cardboard along the dotted lines to form an open rectangular box with base ABCD and height x cm. Find the largest possible volume of the box.
Solution:
Let x be the length of the side of the square cut from each corner of the sheet of cardboard then the height of the box is x. Find the length of the box. key in your answer.
Find the breadth of the box key in your answer.
Volume of the box is equal to length into breadth into height. Substituting the Value of lb and h we get volume is equal to 4 x cube minus 26 x square plus 40 x. We differentiate the equation Vx is equal to 4 x cube minus 26 x square plus 40 x key in the values we get v dash x is equal to 12 x square minus 52 x plus 40.
We now take the second derivative key in the values we get v 2dash x is equal to 24x minus 52.
We take the equation for v dash x and factorize it key in the values we get v dash x is equal to 4 into x minus 1 into 3x minus 10, v dash x is equal to 0 gives x is equal to 1 or 10 by 3. But x is not equal to 10 by 3 because the value x equal to 10 by 3 gives a negative value for the breadth. So x is equal to 1.
we use the second derivative test and find v 2 dash 1 by substituting x is equal to 1. In the equation for V 2 dash x V 2 dash 1 is less than 0.
Therefore, x is equal to 1is the point of maxima. we substitute x is equal to 1 in V x
We get V 1 is equal to 18. If we are remove a square of side 1 cm from each corner of the sheet and make a box of the largest possible volume of the box is 18 cubic cm.
The problem can be picturize with a graph. Observe the graph carefully. From the graph we find that the maximum volume of the box is equal to 18 cubic cm.
Problem on Maxima and Minima
A sheet of cardboard 8 cm by 5 cm is made into a box by cutting equal sized squares from each corner and folding the cardboard along the dotted lines to form an open rectangular box with base ABCD and height x cm. Find the largest possible volume of the box.
Solution:
Let x be the length of the side of the square cut from each corner of the sheet of cardboard then the height of the box is x. Find the length of the box. key in your answer.
Find the breadth of the box key in your answer.
Volume of the box is equal to length into breadth into height. Substituting the Value of lb and h we get volume is equal to 4 x cube minus 26 x square plus 40 x. We differentiate the equation Vx is equal to 4 x cube minus 26 x square plus 40 x key in the values we get v dash x is equal to 12 x square minus 52 x plus 40.
We now take the second derivative key in the values we get v 2dash x is equal to 24x minus 52.
We take the equation for v dash x and factorize it key in the values we get v dash x is equal to 4 into x minus 1 into 3x minus 10, v dash x is equal to 0 gives x is equal to 1 or 10 by 3. But x is not equal to 10 by 3 because the value x equal to 10 by 3 gives a negative value for the breadth. So x is equal to 1.
we use the second derivative test and find v 2 dash 1 by substituting x is equal to 1. In the equation for V 2 dash x V 2 dash 1 is less than 0.
Therefore, x is equal to 1is the point of maxima. we substitute x is equal to 1 in V x
We get V 1 is equal to 18. If we are remove a square of side 1 cm from each corner of the sheet and make a box of the largest possible volume of the box is 18 cubic cm.
The problem can be picturize with a graph. Observe the graph carefully. From the graph we find that the maximum volume of the box is equal to 18 cubic cm.
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