Question

Help me with the workings to these questions please!!!

Question 4 d when a 8.9, b 13.0 and c 7.8 will equal (correct to four decimal places) Not answered Marked out of 1.00 Flag question 4cr Answer: The correct answer is:0.0464 Question 5 Not answered Marked out of 1.00 ch For 3ry 2v-4,find at x 4.0 andy 4.3 (correct to four decimal places) Flag question Answer: The correct answer is: -1.6145 Question 6 Not answered Marked out of 1.00 units are produced and b 66.24 is (correct to two decimal places) Demand for a product is given by the function p -b-48 where p ls price and q is quantty. The value of Marginal Revenue when 66 Flag question Answer: The correct answer is: -1878.84

Answer 1

1. $\large \int_{a}^{b}\frac{1}{4+cx}dx$

Where a= 8.9, b = 13 and c= 7.8

Let us assume 4+7.8x = u

Now differentiate u wrt x we get

$\large \frac{du}{dx}= \frac{d(4+7.8x)}{dx}$

$\large \frac{du}{dx}= 7.8$

$\large \frac{du}{7.8}= dx$

When a= 8.9 then u = 4 + 7.8*8.9 = 73.42

When b = 13, u = 4+ 7.8*13 = 105.4

Now plug in this into the main equation.

$\large \int_{a}^{b}\frac{1}{4+cx}dx$

= $\large \int_{73.42}^{105.4}\frac{1}{7.8u}du$

=$\large \frac{1}{7.8}*\int_{73.42}^{105.4}\frac{1}{u}du$

=$\large \frac{1}{7.8}*ln\left | u \right |_{73.42}^{105.4}$

= (1/7.8)*[ln(105.4) -ln(73.42)]

=(1/7.8) *(4.657762 - 4.296196)

=(1/7.8)*0.361566

= 0.046354

=0.0464 (Approximately)

2. Given, $\large 3x^3y^2 = 2y-4$

Differentiating above equation wrt x we get

$\large \frac{d(3x^3y^2 )}{dx}= \frac{d(2y-4)}{dx}$

$\large 9x^2y^2 + 6x^3y\frac{dy}{dx}=2 \frac{dy}{dx}$

$\large 9x^2y^2 =(2- 6x^3y)\frac{dy}{dx}$

$\large \frac{dy}{dx} = \frac{9x^2y^2}{2- 6x^3y}$

Plug in the value of x and y

$\large \frac{dy}{dx} = \frac{9*4^2*4.3^2}{2- 6*4^3*4.3}$

$\large \frac{dy}{dx} = \frac{2662.56}{2- 1651.2}$

$\large \frac{dy}{dx} =- \frac{2662.56}{1649.2}$

$\large \frac{dy}{dx} = - 1.614455$

Approximately the answer is -1.6145

3. Given $\large P = b - 48q^{\frac{3}{4}}$

Total Revenue = PQ

$\large TR = Pq =( b - 48q^{\frac{3}{4}})q$

$\large TR = bq - 48q^{(1+\frac{3}{4})}$

$\large TR = bq - 48q^{\frac{7}{4}}$

Differentiating above equation wrt q we get

$\large MR = b - 48*\frac{7}{4}*q^{\frac{3}{4}}$

Now plug in b = 66.24, q = 66

$\large MR = 66.24 - 48*\frac{7}{4}*66^{\frac{3}{4}}$

MR = - 1878.84

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