does anyone know how to do this? (1 Viewer)

[lolzlolz]

 

[icycledough]

With part a, it requires a bit of manipulation and converting everything into x terms.

So if we know xy = 400 (area of poster), then y = 400/x

With photograph --> width = x - 10 and length = y - 10 = 400/x - 10

With the dimensions as x - 10 and 400/x - 10, you just multiply and should get the equation given above



With part b, to find the maximum, we need to find the x-value for dA/dx = 0.

Differentiating the equation gives us dA/dx = 4000/(x^2) - 10

Then you solve dA/dx = 0 (also good to prove it is a maximum by finding the 2nd derivative and proving it is negative; negative = concave down = maximum); should be x = 20

Then sub in x = 20 back into the dimensions (x - 10) and (400/x - 10) to find the area, and it should be 10 * 10 = 100cm^2

 

[lolzlolz]

With part a, it requires a bit of manipulation and converting everything into x terms.

So if we know xy = 400 (area of poster), then y = 400/x

With photograph --> width = x - 10 and length = y - 10 = 400/x - 10

With the dimensions as x - 10 and 400/x - 10, you just multiply and should get the equation given above



With part b, to find the maximum, we need to find the x-value for dA/dx = 0.

Differentiating the equation gives us dA/dx = 4000/(x^2) - 10

Then you solve dA/dx = 0 (also good to prove it is a maximum by finding the 2nd derivative and proving it is negative; negative = concave down = maximum); should be x = 20

Then sub in x = 20 back into the dimensions (x - 10) and (400/x - 10) to find the area, and it should be 10 * 10 = 100cm^2

why is the width 10 - x

 

[5uckerberg]

why is the width 10 - x

You mean x-10 because there are two margins and each margin is 5cm. Double that and you will have 10cm.

 

[shadowlike04]

a)
xy = 400
y= 400/x

Area of the photograph is: (x-10)(y-10) = xy -10x +10y +100

Sub in xy value and y value: 400 -10x +10(400/x) +100
=500 -10x +4000/x which is the equation above.

b)

Differentiate the equation and solve for A=0. Sub x value into original equation that is max area.
Check if it is a max with second derivative, should be <0.