Optimization Problems 1. Find two positive numbers such that...

Optimization Problems 1. Find two positive numbers such that the product is 192 and the sum of the first plus three times the second is minimized. a) 32,6 b) 24, 8 c) 16,12 d) 8,24 2. A rectangle is inscribed in an isosceles triangle of height 10 and base 5. Find the dimension of the bottom of the rectangle so that area of the rectangle is maximized. (Use similar triangles.) 10 5 a) 5/3 b) 3 c) 2 d) 5/2 3. Find the x-value of the point on curve y = ?x - 2 that is closest to (11,0). a) 23/2 b) 21/2 c) 10 d) 19/2 4. An open-top box is created by folding a piece of 2 by 3 feet cardboard with four equal sized squares cut out of the corners. Find x, the side of one cut-out square so that the volume is maximized. (Use the quadratic formula; only one of the solutions makes sense given the dimensions of the cardboard.) a) 1.27 b) .5 c) d) 0.4

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00:01 In the first question, let there be two positive numbers as a and b as we are already given that product of a and b is 132 and so b is equal to 132 divided by a and we are given that a plus 3b is equal to p from which we can say that by putting the value of b we can say that a plus 3 times 132 divided by a on derivating p with respect to a we get 1 plus 576 divided by a square when we put this is equal to 0 divided by a square is equal to 0 we get the value of a square is equal to 576 from which we get a is equal to 24 and from above equation we get 1 b is equal to 132 divided by 24 which is equal to 8 so a is equal to 24 and b is equal to 8 this is our answer for the first part now moving on to the second question we have in question we have diagram as we have diagram such that we will mark the points as a b and c and edges as a a as it is an isosceles triangle so both sides are equal and the width of the rectangle as y and length as x as we know that area of isosceles triangle is b by b divided by 2 under root a square minus b square by 4 where b is our base and a is length of one side of triangle which is equal to 1 divided by 2 multiplied with 5 multiplied with 10 where 10 is the height and 5 is the base moving on we have b divided by 2 raised to the power 2 multiplied with a raised to the power 2 minus b b square divided by 4 which is equal to 25 raised to the power 2 from which we get b square divided by 4 multiplied with a square minus b square divided by 4 is equal to 625 which gives 25 divided by 4 as the base is 5 multiplied with a square minus 25 divided by 4 is equal to 625 which can be further simplified as a square is equal to 100 plus 25 divided by 4 which is is equal to 425 divided by 4 which gives the value of a as 5 under root 17 divided by now for sine of b and sine of c as we know that 1 divided by 2 multiplied with 5 multiplied with 5 under root 17 divided by 4 multiplied with sine of c is equal to 25 as a is equal to 1 area is equal to 1 divided by 2 a multiply with a multiply with b multiply with sine of theta sine of either one of the angle of the side either one of the angle thus the side of the triangle making with the base from which we get sine c is equal to 4 divided by 17 which gives 10 c is equal to 4 which is also equal to 10 b now area of rectangle rectangle in is given by a is equal to x multiply with 5 minus x by 10 of b minus x by 10 of c these are all these which gives us x multiply with 5 minus x divided by 4 minus x divided by 4 which gives x multiplied with 5 minus x by 2 on simplifying we get on derivating d a with respect to x we get d a divided by d x is equal to 5 minus 2 x by 2 and on putting it is and putting it is equal to 0 we get x is equal to 5 and from which we get area is equal to 5 multiply with 5 minus 5 divided by 2 which is equal to 25 divided by 2 hence the bottom dimension bottom dimension is 5 minus x divided by 10 of b minus x divided by 10 of c which is equal to 5 minus 5 divided by 4 minus 5 divided by 4 which gives 5 minus 5 divided by 2 which is is equal to 5 divided by 2 so option now moving on to the third question we have according to the question we have y is equal to x minus 2 which gives y square is equal to x minus from which we get the graph as we get the graph as this in which we got 2 comma 0 if we put y is equal to 0 and y comma 0 is a point from who between which the distance we have we need to find so moving forward let a b is a point let a b is a point from which the distance we need to find from y comma 0 so the distance is equal to 11 minus a raised to the power 2 plus b square whole under root and as we know b is equal to under root a minus 2 which we get from the equation now the distance can be written as 1 to 1 plus a square plus b square minus 22 a whole under root which can be simplified as d 121 plus a square plus a minus 2 minus 22 a from first we get this from first now moving forward the distance is equal to a square minus 21 a plus 11 a plus 119 sorry this is 119 plus 119 from which by differentiating the distance d with respect to a we get d d divided by d a is equal to 1 by 2 multiplied with a square minus 21 a plus 119 raised to the power minus 1 by 2 from which we get 2 a minus 21 is equal to 0 from which we get a is equal to 21 divided by 2 so b is equal to under root 21 minus 4 divided by 2 which is equal to 17 divided by 2 now the distance we get from this is equal to 11 minus 21 divided by 2 raised to the power 2 plus 17 divided by 2 and d is equal to 1 divided by 4 plus 17 divided by 2 is equal to 35 divided by 2 this is our distance so x is equal to 21 divided by option 3...

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