Answer:
the dimensions of the poster with the smallest area is 36cm by 42cm
Step by step Explanation:
Let us denote the width and height of the printed area as x and y respectively.
Then xy=864cm^2
y= 864/x
The total height of the poster , along with 9cm margin at sides is (y+18)
The total width of the poster , along with 6cm margin at sides is (y+12)
the area of the total poster is
A= (x+12)(y+18)...............eqn(1)
Substitute the value of y into equation (1)
A= (x+12)[(⁸⁶⁴/ₓ)+(18)]
864+ 18x +¹⁰³⁶⁸/x +216
18x +( ⁰³⁶⁸/x) + 1080
then find the derivative of x in this equation by
A'= 18-(¹⁰³⁶⁸/x²)
then equate A' to be zero so as to find the value of x
0= 18- (¹⁰³⁶⁸/x²)
18= (¹⁰³⁶⁸/x²)
x²= 576
x= √576
x= 24
Since A'' is positive for x> 0 then A is concave up and x= 24 I s the minimum value, the value of y as well is
y= 864/x
y= 432
Then the width of the poster is now
x+12= 24+ 12= 36
The corresponding height is
y+8= 24+18= 42
Therefore,the dimensions of the poster with the smallest area is 36cm by 42cm
