A wire of length 36 cm is cut into two pieces. One of the pieces

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 Multiple Choice QuestionsShort Answer Type

301.

Manufacturer can sell x items at a price of rupees open parentheses 5 minus straight x over 100 close parentheses space each. The cost price of x items is Rs. open parentheses straight x over 5 plus 500 close parentheses. Find the number of items he should sell to earn maximum profit. 

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302.

A firm has found from experience that its profit as a function of x, the amount produced, is given by
straight p left parenthesis straight x right parenthesis space equals space minus straight x cubed over 3 plus 729 space straight x space minus 2500 comma space space space 0 space less or equal than space straight x space less or equal than space 35.
Find the value of x that maximises the profit.

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303.

A beam of length l is supported at one end. If W is the uniform load per unit length, the bending moment M at a distance x from the end is given by straight M space equals 1 half space straight l space straight x space space minus space 1 half space straight W space straight x squared. maximum value. Find the point on the beam at which the bending moment has th

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 Multiple Choice QuestionsLong Answer Type

304. Given the sum of the perimeters of a square and a circle, prove that the sum of their areas is least when the side of the square is equal to the diameter of the circle.
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305.

A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.


Total length of wire = 36 cm
Let x cm be length of each side of square and y cm be length of each side of quilateral triangle.
Length of wire used for square = 4x cms
and length of wire used for triangle = 3y cms
∴  4x + 3y = 36  ⇒  3y = 36 – 4x
rightwards double arrow space space space space straight y space equals space fraction numerator 36 minus 4 straight x over denominator 3 end fraction                                                                      ...(1)
Let A denote the sum of the areas of square and equilateral triangle.
therefore space space space straight A space equals space left parenthesis straight x right parenthesis squared plus 1 half left parenthesis straight y right parenthesis space left parenthesis straight y right parenthesis space left parenthesis sin space 60 degree right parenthesis                                    open square brackets increment space equals space 1 half bc space sin space straight A close square brackets

             straight x squared plus fraction numerator square root of 3 over denominator 4 end fraction straight y squared space equals space straight x squared plus fraction numerator square root of 3 over denominator 4 end fraction open parentheses fraction numerator 36 minus 4 straight x over denominator 3 end fraction close parentheses squared                   open square brackets because space of space left parenthesis 1 right parenthesis close square brackets 
therefore space space space space space space straight A minus straight x squared plus fraction numerator 4 square root of 3 over denominator 9 end fraction left parenthesis 9 minus straight x squared right parenthesis
                dA over dx equals 2 straight x plus fraction numerator 4 square root of 3 over denominator 9 end fraction.2 left parenthesis 9 minus straight x right parenthesis thin space left parenthesis negative 1 right parenthesis space equals space 2 straight x minus fraction numerator 8 square root of 3 over denominator 9 end fraction left parenthesis 9 minus straight x right parenthesis
dA over dx space equals space 0 space space space space gives space us space space 2 straight x minus fraction numerator 8 square root of 3 over denominator 9 end fraction left parenthesis 9 minus straight x right parenthesis space equals space 0

therefore space space space straight x minus 4 square root of 3 space plus space fraction numerator 4 square root of 3 over denominator 9 end fraction straight x space equals space 0 space space space space rightwards double arrow space space space space open parentheses 1 plus fraction numerator 4 square root of 3 over denominator 9 end fraction close parentheses straight x space equals space 4 square root of 3
rightwards double arrow space space space space space open parentheses 9 plus 4 square root of 3 close parentheses straight x space equals space 36 square root of 3 space space space rightwards double arrow space space space straight x space equals space fraction numerator 36 square root of 3 over denominator 9 plus 4 square root of 3 end fraction
space space space space space space space space space space space fraction numerator straight d squared straight A over denominator dx squared end fraction space equals space 2 plus fraction numerator 8 square root of 3 over denominator 9 end fraction
At space space straight x equals space fraction numerator 36 square root of 3 over denominator 9 plus 4 square root of 3 end fraction comma space space space space fraction numerator straight d squared straight A over denominator dx squared end fraction space equals space 2 plus fraction numerator 8 square root of 3 over denominator 9 end fraction greater than 0
therefore space space space straight A space is space minimum space when space straight x space equals space fraction numerator 36 square root of 3 over denominator 9 plus 4 square root of 3 end fraction
         Length of piece required for square  = 4x = fraction numerator 144 square root of 3 over denominator 9 plus 4 square root of 3 end fraction cm
and length of piece required for triangle  = 3y = 36 - 4x
               equals space 36 minus fraction numerator 144 square root of 3 over denominator 9 plus 4 square root of 3 end fraction space equals space fraction numerator 324 plus 144 square root of 3 minus 144 square root of 3 over denominator 9 plus 4 square root of 3 end fraction space equals space fraction numerator 324 over denominator 9 plus 4 square root of 3 end fraction cm.


      

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306. A figure consists of a semi-circle with a rectangle on its diameter. Given perimeter of the figure, find the dimensions in order that the area may be maximum.
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307.

A window consists of a semi-circle with a rectangle on its diameter. If the perimeter of the window is 30 metres, find the dimensions of the window in order that its area may be maximum.
Or
A window is in the form of a rectangle surmounted by a semi-circle. If the total perimeter of the window is 30 m. find the dimensions of the window so that maximum light is admitted.

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308. A window is in the form of a rectangle surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
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309.

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

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 Multiple Choice QuestionsShort Answer Type

310. A wire of length 25 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What would be the lengths of the two pieces, so,that combined area of the square and the circle is minimum?
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