The equation in y=mx+b form is: Y=2x+10
X^2 + y^2 = (3x^2 + 2y^2 - x)^2
2x + 2y f'(x) = 2(3x^2 + 2y^2 - x)(6x + 4y f'(x) - 1) = 36x^3 + 24x^2yf'(x) + 24xy^2 + 16y^3f'(x) - 4y^2 - 18x^2 - 8xyf'(x) + x
f'(x)(2y - 24x^2y - 16y^3 + 8xy) = 36x^3 + 24xy^2 - 4y^2 - 18x^2 - x
f'(x) = (36x^3 + 24xy^2 - 4y^2 - 18x^2 - x)/(2y - 24x^2y - 16y^3 + 8xy)
f'(0, 0.5) = -4(0.5)^2/(2(0.5) - 16(0.5)^3) = -1/(1 - 2) = -1/-1 = 1
Let the required equation be y = mx + c; where y = 0.5, m = 1, x = 0
0.5 = 1(0) + c = 0 + c
c = 0.5
Therefore, the tangent line at point (0, 0.5) is
y = x + 0.5
Answer:
ef nc h hhjkev b kSDNS m c fm dsj hv hj nd,z,hjzhjchj n nn ,hjk sdm s d kjlj, nvdshbh fan vmnxjhsadj nf nc yguxcugyaefmnMNJLUIUIGCXHJ NNRJK IERJK MNMNCJVNUI EFA JKAINIP NN VJKEUIOAU
Step-by-step explanation:
so sorry
Answer:
<h2> $9</h2>
Step-by-step explanation:
<h2>
f(x) = - 25 ( x - 9 )² + 200</h2>
<em>when</em><em> </em>x = 5 ,
f(5) = - 25 ( 5 - 9 )² + 200 = - 200
<em>when</em><em> </em>x = 9 ,
f(9) = - 25 ( 9 - 9 )² + 200 = 200
<em>when </em>x = 10 ,
f(10) = - 25 ( 10 - 9 )² + 200 = 175
<em>when</em><em> </em>x = 15 ,
f(15) = - 25 ( 15 - 9 )² + 200 = -700
∴ when <u>price = $9</u>
the company receive the maximum profit