QQ - Calculus (1 Viewer)

[Smile12345]

Hello All...

Can you please help me with the following question?

A poster consists of a photograph bordered by a 5cm margin. The area of the poster is to be .

I know the area of the photograph is given by the equation .

But how do I find the maximum area possible for the photograph??

Thanks ;D

 

[strawberrye]

Follow these steps
1)differentiate the area formula, and let it equal to zero to find possible values of x-remember because x is a length, hence if it is negative or zero, you can reject those values,
2)determine the nature of the x values you just find-by finding the second derivative, and substituting the values of x into the second derivative respectively, if it is less than 0-you can confirm for that value of x-that is the maximum area
3)then just sub that value of x into A-and ta da, there you are, the maximum area possible-make sure you write it with the correct units as well

 

[bokat]

Write A as below:
A=500-10x-4000x^-1
Differentiate:
dA/dx=-10+4000x^-2
Equate this to zero.
-10+4000x^-2=0 and solve to get:
x=20 We do not take x=-20 as x is length and cannot be negative.
Find the second derivative to check to see if this value of x gives the max value:
A''=-8000x^-3 Now substitute x=20 in this to get:
A''=-800(20)^-3 This is a negative value so x=20 gives the max value for A.
Sub this value in A to find tht max value:
A=500-10(20)-4000/(20)=100 square cm

Hope it helps,
Reza Bokat,
Cambridge Coaching
www.cambridgecoaching.com.au