A Course of Elementary Mathematics Affording Aid to Candidates for Admission I
A Course of Elementary Mathematics Affording Aid to Candidates for Admission I
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In our eReader you can find the full English version of the book. Read A Course of Elementary Mathematics Affording Aid to Candidates for Admission I Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book A Course of Elementary Mathematics Affording Aid to Candidates for Admission I
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To quickly assess the difficulty of the text, read a short excerpt:
And a min. , it is necessary to notice whether or not the sign oi f^(x) has been changed when equating it to zero : the expression represented by u above having been transposed, the sign of it has been changed. [In next prob. The sign will be pre- served. ] 284 SCHOLIUM.
Peob. 9. To determine when the area of the part of the leaf turned down, as in last problem, is the least possible.
It is plain that B^B is bisected at right angles by PQ, therefore PQ . B'B=A: times the area of B'PQ. Now by la...st prob. , PQz=. /- and B'B=.^2ax, . '. Multiplying the squares of ttese together, we have V 2iXa '7* 2 y* ~ CL 4a- =min. , . -. ;^=max. =M, . '. /{x)=2xaux*=0, . : f^{x)=2 iux^=0.
2xa X 3 2a Multiply this by -, and subtract, . *. -cc a=0, . -. X=. , and since /2(a7)= 12m^^ is negative, m, for this value of x^ is a maximum ; and therefore the proposed expression is a minimum.
Examples for Exercise.
(1) Of all right-angled plane triangles baring the same hypotenuse, to find that which has the greatest area.
What to read after A Course of Elementary Mathematics Affording Aid to Candidates for Admission I?
You can find similar books in the "Read Also" column, or choose other free books by J R John Radford Young to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
And a min. , it is necessary to notice whether or not the sign oi f^(x) has been changed when equating it to zero : the expression represented by u above having been transposed, the sign of it has been changed. [In next prob. The sign will be pre- served. ] 284 SCHOLIUM.
Peob. 9. To determine when the area of the part of the leaf turned down, as in last problem, is the least possible.
It is plain that B^B is bisected at right angles by PQ, therefore PQ . B'B=A: times the area of B'PQ. Now by la...st prob. , PQz=. /- and B'B=.^2ax, . '. Multiplying the squares of ttese together, we have V 2iXa '7* 2 y* ~ CL 4a- =min. , . -. ;^=max. =M, . '. /{x)=2xaux*=0, . : f^{x)=2 iux^=0.
2xa X 3 2a Multiply this by -, and subtract, . *. -cc a=0, . -. X=. , and since /2(a7)= 12m^^ is negative, m, for this value of x^ is a maximum ; and therefore the proposed expression is a minimum.
Examples for Exercise.
(1) Of all right-angled plane triangles baring the same hypotenuse, to find that which has the greatest area.
What to read after A Course of Elementary Mathematics Affording Aid to Candidates for Admission I?
You can find similar books in the "Read Also" column, or choose other free books by J R John Radford Young to read onlineMoreLess
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