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Calculate-the-perimeter-of-a-rectangle-whose-area-is-represented-by-the-polynomial-25x-2-35x-12-Given-that-the-length-and-breath-are-not-constant-

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Question Number 220577 by fantastic last updated on 16/May/25
Calculate the perimeter of a rectangle whose area is represented by the polynomial 25x^2 −35x+12(Given that the length and breath are not constant)
$${Calculate}\:{the}\:{perimeter}\:{of}\:{a}\:{rectangle} \\ $$$${whose}\:{area}\:{is}\:{represented}\:{by}\:{the}\:{polynomial} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}\left({Given}\:{that}\:{the}\:{length}\:{and}\:{breath}\:{are}\:{not}\:{constant}\right) \\ $$
Answered by Frix last updated on 16/May/25
A=ab=25x^2 −35x+12 ⇒ b=((25x^2 −35x+12)/a) ⇒ P=2(a+b)=((2(a^2 +25x^2 −35x+12))/a) A>0∧P>0 ⇒ a>0∧b>0∧(x<(3/5)∨(4/5)<x) We can do this: Let x=t+(4/5) ⇒ because the polynomial is symmetric we can let t>0 A=25x^2 −35x+12=5t(5t+1) P=((2(a^2 +5t(5t+1)))/a); a>0∧t>0
$${A}={ab}=\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}\:\Rightarrow\:{b}=\frac{\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}}{{a}}\:\Rightarrow \\ $$$${P}=\mathrm{2}\left({a}+{b}\right)=\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}\right)}{{a}} \\ $$$${A}>\mathrm{0}\wedge{P}>\mathrm{0}\:\Rightarrow\:{a}>\mathrm{0}\wedge{b}>\mathrm{0}\wedge\left({x}<\frac{\mathrm{3}}{\mathrm{5}}\vee\frac{\mathrm{4}}{\mathrm{5}}<{x}\right) \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{do}\:\mathrm{this}: \\ $$$$\mathrm{Let}\:{x}={t}+\frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow\:\mathrm{because}\:\mathrm{the}\:\mathrm{polynomial}\:\mathrm{is} \\ $$$$\mathrm{symmetric}\:\mathrm{we}\:\mathrm{can}\:\mathrm{let}\:{t}>\mathrm{0} \\ $$$${A}=\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}=\mathrm{5}{t}\left(\mathrm{5}{t}+\mathrm{1}\right) \\ $$$${P}=\frac{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{5}{t}\left(\mathrm{5}{t}+\mathrm{1}\right)\right)}{{a}};\:{a}>\mathrm{0}\wedge{t}>\mathrm{0} \\ $$
Commented by fantastic last updated on 16/May/25
Do not think with so much complexity
$$\:{Do}\:{not}\:{think}\:{with}\:{so}\:{much}\:{complexity} \\ $$
Commented by Frix last updated on 16/May/25
But that′s what you asked for. Without any other conditions given, this is the answer.
$$\mathrm{But}\:\mathrm{that}'\mathrm{s}\:\mathrm{what}\:\mathrm{you}\:\mathrm{asked}\:\mathrm{for}.\:\mathrm{Without}\:\mathrm{any} \\ $$$$\mathrm{other}\:\mathrm{conditions}\:\mathrm{given},\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}. \\ $$
Commented by Frix last updated on 16/May/25
But it′s wrong. ab=25x^2 −35x+12 We are free to choose the values of 2 variables. Example Let x=3∧a=4 ⇒ b=33 2(a+b)=74 But 20x−14=46
$$\mathrm{But}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}. \\ $$$${ab}=\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12} \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{2} \\ $$$$\mathrm{variables}. \\ $$$$\mathrm{Example} \\ $$$$\mathrm{Let}\:{x}=\mathrm{3}\wedge{a}=\mathrm{4}\:\Rightarrow\:{b}=\mathrm{33} \\ $$$$\mathrm{2}\left({a}+{b}\right)=\mathrm{74} \\ $$$$\mathrm{But}\:\mathrm{20}{x}−\mathrm{14}=\mathrm{46} \\ $$
Commented by mr W last updated on 16/May/25
the question is just like to ask: the product of the ages of a couple is x^2 +3x−5, what are their weights?
$${the}\:{question}\:{is}\:{just}\:{like}\:{to}\:{ask}: \\ $$$${the}\:{product}\:{of}\:{the}\:{ages}\:{of}\:{a}\:{couple}\:{is} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{5},\:{what}\:{are}\:{their}\:{weights}? \\ $$
Commented by fantastic last updated on 16/May/25
Sir i also know it it came in our school exam
$${Sir}\:{i}\:{also}\:{know}\:{it}\:{it}\:{came}\:{in}\:{our}\:{school}\:{exam} \\ $$
Commented by fantastic last updated on 16/May/25
Commented by fantastic last updated on 16/May/25
Q.4.1
$${Q}.\mathrm{4}.\mathrm{1} \\ $$
Commented by Ghisom last updated on 16/May/25
I think on this level the answer should be: A=25x^2 −35x+12=(5x−4)(5x−3) ⇒ a=5x−3∧b=5x−4 P=2(a+b)=20x−14
$$\mathrm{I}\:\mathrm{think}\:\mathrm{on}\:\mathrm{this}\:\mathrm{level}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}: \\ $$$${A}=\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}=\left(\mathrm{5}{x}−\mathrm{4}\right)\left(\mathrm{5}{x}−\mathrm{3}\right)\:\Rightarrow \\ $$$${a}=\mathrm{5}{x}−\mathrm{3}\wedge{b}=\mathrm{5}{x}−\mathrm{4} \\ $$$${P}=\mathrm{2}\left({a}+{b}\right)=\mathrm{20}{x}−\mathrm{14} \\ $$
Commented by Ghisom last updated on 16/May/25
I know it′s wrong. weird to teach false mathematics...
$$\mathrm{I}\:\mathrm{know}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}.\:\mathrm{weird}\:\mathrm{to}\:\mathrm{teach}\:\mathrm{false} \\ $$$$\mathrm{mathematics}… \\ $$
Commented by fantastic last updated on 16/May/25
...
$$… \\ $$

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