What is 20-23 (work does not need to be shown)
1 answer:
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Answer:
19 by 38
Step-by-step explanation:
The area of the poster, A=xy
Margin at both up and bottom is 2inches, at the right and left is 1 inch
Then we can say
(x-4)(y-2)= 578
Let us make y subject of the formula
(y-2)= 578/(x-4)
y=[ 578/(x-4)] + 2
If we substitute into the area A equation we have
A= x { [ 578/(x-4)] + 2}
A= 2x + (578x)/(x - 4)
If we differentiate we have
A'(x)= [2+578(x-4)+578x ]/ (x-4)^2
=[ 2(x-4)^2 + 578(x -4) + 578x ] / (x-4)^2
=[ 2(x^2 - 8x +16) + 578(x-4) + 578x ] /(x-4)^2
If we simplify this we have
=[ 2x^2 -16x +32+578x -578x - 2312]/ (x-4)^2
=( 2x^2 -16x - 2312) / (x-4)^2
At A'(x)= 0
= ( x^2 - 8x - 2312) / (x-4)^ =0
If we divide through by 2 we have
x^2 - 8x - 1156= 0
Solving the quadratic eqn( CHECK THE ATTACHMENT)
X= 38 or -30 then we choose the positive one, then x= 38
Then from area of the poster
A= (x-4)(y-2)= 578
Substitute 38 as value of x
(38-4)(y-2)= 578
34(y-2)=578
34y-68=578
34y=646
y=19
Hence dimensions of the poster is 19 by 38
CHECK THE ATTACHMENT FOR QUADRATIC SOLUTION
