Help, picture attached. Show working thanks I only need help with 1.d) which is finding the domain of A(x)
what's your area function?
your area should be a quadratic (upside down parabola) since area is positive, set your area function = 0. you'll get the two x-values that make your area zero. the x's between these two values is your domain of A(x).
yep but how do u solve it? 0=10y + xy -x^2
What do you mean, how do you solve it? You said you need the domain?
yeh how do u solve 0=10y + xy - x^2?? @dpaInc said i need to solve it first to figure out that domain.
If you're asking how to solve for y, you add x^2 to both sides so you get x^2=10y+xy Pull a y out front of the right side and you get x^2=y(10+x) And therefore y=x^2/(10+x)
your area is in terms of two variables. get it in terms of 1 by using the perimeter constraint.
Is this for a Calculus class?
so i can use the equation from part c?
nope, i haven't learnt Calculus yet.
domain of y=x^2/(10+x) \[{x \in \mathbb{R}~:x+10 \neq 0}\]
yes @JayDS
um i solved it but i get the wrong x values?
\[0=2x ^{2}+16x+260\] \[-260 = -2x ^{2} + 16x\] \[-260 -2x(x-8)\] \[-2x = -26\] or x-8=-260 x = 130 or x = -252
-2x = -260*
your area function should be A(x) = -2x^2 + 16x + 260
yep can u solve it? when i solved it i seemed to have got the wrong answers as shown in my working above.
setting A(x)=0, i get irrational roots...
hmm i'm not too sure but the answers say the domain is (0, 13) and i'm not sure how they got that
hang on.. my perimeter equation was wrong...
the P equation is y = -x + 26
perimeter*
hmmm i keep getting the same thing... my perimeter is the same
wait.. look at you perimeter equation... x can only be as big as y.
it doesn't make sense if x is bigger than y (according to the figure).
so the most x can get is 13... the least is 0.
um i don't quite get it
the Perimeter equation is y= - x + 26 but how can you tell that x at most is 13? u can't.
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